FP1 COMPLEX NUMBERS PAST EXAM QUESTIONS Nos.1-7 Are

FP1 COMPLEX NUMBERS PAST EXAM QUESTIONS

Nos.1-7 are standard plain-vanilla questions.

Questions 8, 9, 10, 11, 13, 19, 26, 27, 30 demand a bit more thought.

Q.21 is very easy if you use the sum-of-roots/ product-of-roots method, but a bit tricky otherwise.

For Q.10 and Q.30, remember that if 64 is thought of as a complex number, then its argument is 0 or 2π or 4π or... When you multiply complex numbers, you add arguments. Therefore, when you take powers of complex numbers, you multiply arguments. When you take roots of complex numbers, you divide arguments. Therefore, the cube roots of 64 all have modulus 4, and they have arguments 0, 2π/3, 4π/3.

1. Given that z = –3 + 4i,

(a) find the modulus of z, (2)

(b) the argument of z in radians to 2 decimal places. (2)

– 14  2i Given also that w = z

The complex numbers z and w are represented by points A and B on an Argand diagram.

(d) Show the points A and B on an Argand diagram. (2) (Total 10 marks)

2. f(x) = 2x3 – 5x2 + px – 5, p 

Given that 1 – 2i is a complex solution of f (x) = 0,

(a) write down the other complex solution of f(x) = 0, (1)

(b) solve the equation f(x) = 0, (6)

3. z = 2 – 3i

(a) Show that z2 = –5 – 12i. (2)

Find, showing your working,

(b) the value of z 2 , (2)

City of London Academy 1

(d) Show z and z2 on a single Argand diagram. (1) (Total 7 marks)

4. f(x) = x3 + x2 + 44x + 150

Given that f(x) = (x + 3)(x2 + ax + b), where a and b are real constants,

(a) find the value of a and the value of b. (2)

(b) Find the three roots of f(x) = 0. (4)

5. The complex numbers z1 and z2 are given by

z1= 2 + 8i and z2 = 1 – i

Find, showing your working,

z (a) 1 in the form a + bi, where a and b are real, (3) z 2

z (b) the value of 1 , z 2 (2)

z (c) the value of arg 1 , giving your answer in radians to 2 decimal places. z 2 (2) (Total 7 marks)

City of London Academy 2 6. Given that 2 and 5 + 2i are roots of the equation

x3 – 12x3 + cx + d = 0, c, d  ,

(a) write down the other complex root of the equation. (1)

(b) Find the value of c and the value of d. (5)

7. The complex numbers z1and z2 are given by

z1 = 2 – i and z2 = – 8 + 9i

(a) Show z1 and z2 on a single Argand diagram. (1)

Find, showing your working,

(b) the value of z1 , (2)

z (d) 2 in the form a + bi, where a and b are real. z1 (3) (Total 8 marks)

8. f (x) = (x2 + 4)(x2 + 8x + 25)

(a) Find the four roots of f (x) = 0. (5)

(b) Find the sum of these four roots. (2) (Total 7 marks)

City of London Academy 3 12 5i 9. Given that z1 = 3 + 2i and z2  , z1

(a) find z2 in the form a + ib, where a and b are real. (2)

(b) Show on an Argand diagram the point P representing z1 and the point Q representing z2. (2)

 (c) Given that O is the origin, show that  POQ = . 2 (2)

The circle passing through the points O, P and Q has centre C. Find

(d) the complex number represented by C, (2)

(e) the exact value of the radius of the circle. (2) (Total 10 marks)

10. (a) Write down the value of the real root of the equation

x3 – 64 = 0. (1)

(b) Find the complex roots of x3 – 64 = 0, giving your answers in the form a + ib, where a and b are real. (4)

11. The complex number z is defined by

City of London Academy 4 a  2i z  , a  , a  0 a  i

1 Given that the real part of z is , find 2

(a) the value of a, (4)

(b) the argument of z, giving your answer in radians to 2 decimal places. (3) (Total 7 marks)

12. z = –2 + i

(a) Express in the form a + ib

1 (i) , z

(ii) z2 (4)

(b) Show that z 2  z  5 2 (2)

(d) Display z and z2 – z on a single Argand diagram. (2) (Total 10 marks)

13. z = √3 – i.

z* is the complex conjugate of z.

z 1 3 (a) Show that   i. z * 2 2 (3)

City of London Academy 5

z (b) Find the value of . z * (2)

z (c) Verify, for z = √3 – i, that arg = arg z – arg z*. z * (4)

z (d) Display z, z* and on a single Argand diagram z * (2)

(e) Find a quadratic equation with roots z and z* in the form ax2 + bx + c = 0, where a, b and c are real constants to be found. (2) (Total 13 marks)

14. (a) Find the roots of the equation

z2 + 2z + 17 = 0,

giving your answers in the form a + ib, where a and b are integers. (3)

(b) Show these roots on an Argand diagram. (1) (Total 4 marks)

15. The complex numbers z1 and z2 are given by

z1 = 5 + 3i,

z2 = 1 + pi

where p is an integer.

z (a) Find 2 in the form a + ib, where a and b are expressed in terms of p. z1

City of London Academy 6 (3)

 z2   Given that arg    ,  z1  4

(b) find the value of p. (2) (Total 5 marks)

16. The complex numbers z and w satisfy the simultaneous equations

2z + iw = –1,

z – w = 3 + 3i.

(a) Use algebra to find z, giving your answer in the form a + ib, where a and b are real. (4)

(b) Calculate arg z, giving your answer in radians to 2 decimal places. (2) (Total 6 marks)

17. Given that 3 – 2i is a solution of the equation

x4– 6x3 + 19x2– 36x + 78 = 0,

(a) solve the equation completely, (7)

(b) show on a single Argand diagram the four points that represent the roots of the equation. (2) (Total 9 marks)

z  2i 18. Given that = i, where  is a positive, real constant, z i

      (a) show that z  1 i 1  2   2  (5)

City of London Academy 7 1 Given also that arg z = arctan , calculate 2

(b) the value of λ, (3)

19. Given that 1 + 3i is a root of the equation z3 + 6z + 20 = 0,

(a) find the other two roots of the equation, (3)

(b) show, on a single Argand diagram, the three points representing the roots of the equation, (1)

20. z = –4 + 6i.

(a) Calculate arg z, giving your answer in radians to 3 decimal places. (2)

A The complex number w is given by w = , where A is a positive constant. Given that 2  i w = 20,

(b) find w in the form a + ib, where a and b are constants, (4)

w (c) calculate arg . z (3) (Total 9 marks)

City of London Academy 8

21. Given that 3 + i is a root of the equation f(x) = 0, where

f(x) = 2x3 + ax2 + bx – 10, a, b  ,

(a) find the other two roots of the equation f(x) = 0, (4)

(b) find the value of a and the value of b. (3) (Total 7 marks)

 3 3  22. Given that z = 4 cos  isin  and w = 1 – i3, find  4 4 

z (a) , w (3)

 z  (b) arg   , in radians as a multiple of .  w  (3)

and 4, respectively.

(3)

(d) Show that  AOC =  DOB. (3)

(e) Find the area of triangle AOC. (2) (Total 14 marks)

23. The complex number z = a + ib, where a and b are real numbers, satisfies the equation

z2 + 16 – 30i = 0.

City of London Academy 9 (a) Show that ab = 15. (2)

(b) Write down a second equation in a and b and hence find the roots of

z2 + 16 – 30i = 0. (4) (Total 6 marks)

w 24. Given that z = 1 + 3i and that = 2 + 2i, find z

(a) w in the form a + ib, where a, b  , (3)

(b) the argument of w, (2)

On an Argand diagram, the point A represents z and the point B represents w.

(d) Draw the Argand diagram, showing the points A and B. (2)

(e) Find the distance AB, giving your answer as a simplified surd. (2) (Total 11 marks)

25. Given that z = 2 – 2i and w = –3 + i,

(a) find the modulus and argument of wz2. (6)

(b) Show on an Argand diagram the points A, B and C which represent z, w and wz2 respectively, and determine the size of angle BOC. (4)

City of London Academy 10 (Total 10 marks)

26. (a) By factorisation, show that two of the roots of the equation x3 – 27 = 0 satisfy the quadratic equation x2 + 3x + 9 = 0. (2)

(b) Hence, or otherwise, find the three cube roots of 27, giving your answers in the form a + ib, where a, b  . (3)

27.

a  3i z = , a  . 2  ai

(a) Given that a = 4, find z. (3)

 (b) Show that there is only one value of a for which arg z = , and find this value. 4 (6) (Total 9 marks)

     2 2  28. z = 4cos  isin  , and w = 3cos  isin  .  4 4   3 3 

Express zw in the form r (cos  + i sin  ), r > 0,  <  < . (Total 3 marks)

29. Given that z = 3 – 3i express, in the form a + ib, where a and b are real numbers,

(a) z2, (2)

1 (b) . z (2)

City of London Academy 11 1 (c) Find the exact value of each of z,z2 and . z (2)

1 The complex numbers z, z2 and are represented by the points A, B and C respectively on an z Argand diagram. The real number 1 is represented by the point D, and O is the origin.

(d) Show the points A, B, C and D on an Argand diagram. (2)

(e) Prove that OAB is similar to OCD. (3) (Total 11 marks)

30.

f(z) = z 3 – 1

(a) Find two complex numbers in the form a + ib, where a, b  , a  0, b  0, for which f(z) = 0. (4)

(b) Display the three roots of the equation f(z) = 0 on an Argand diagram. (1)

1. (a) z  (32  42 )  5 M1A12

4 (b) arg z = π – arctan = 2.21 M1A12 3

– 14  2i (–14  2i)(–3 – 4i) (c) w   M1 – 3  4i (–3  4i)(–3 – 4i)

(42  8)  i(–6  56)  A1A1 9 16

50  50i   2  2i A14 25

(d)

City of London Academy 12 B1B12 [10]

2. (a) 1 + 2i B11

(b) (x – 1 + 2i )(x –1 – 2i) are factors of f(x) M1

so x2 – 2x + 5 is a factor of f (x) M1A1

f(x) = (x2 – 2x + 5)(2x – 1) M1 A1ft

1 Third root is A16 2

3. (a) (2 – 3i)(2 – 3i) = ….. Expand and use i2 = –1, getting completely correct expansion of 3 or 4 terms M1

Reaches – 5 – 12i after completely correct work (must see 4 – 9) (*) A1 cso2

Note

M1: for 4 – 9 – 12i or 4 – 9 – 6i – 6i or 4 – 32 – 12i but must have correct statement seen and see i^2 replaced by –1 maybe later A1: Printed answer. Must see 4 – 9 in working. Jump from 4 – 6i – 6i + 9i^2 to –5 – 12i is M0A0

(b) z 2  (5)2  (12)2 13 or z25 2  12 2  13 M1 A12

Alternative methods for part (b) 2 z 2  z  22 – 32 13 Or: │z2│ = zz* = 13 M1 A12

Note

City of London Academy 13 Method may be implied by correct answer. NB z 2 =169 is M0 A0

12 12 12 5 (c) tanα  ( allow  ) or sin  or cos  M1 5 5 13 13

arg(z 2 )  ( 1.176...)  1.97 (or 4.32) allow awrt A12

Alternative method  5 α = 2 × arctan – 3  (allow 3 ) or use + arctan M1 2 2 2 12

so arg(z2) = – (π – 1.176...) = – 1.97 (or 4.32) allow awrt A1

Note

12 Allow arctan for M1 or    arctan 5 5 2 12

(d)

Both in correct quadrants. Approximate relative scale No labels needed B11 Allow two diagrams if some indication of scale Allow points or arrows [7]

4. (a) a  2, b  50 B1 B12

Note

Accept x2 – 2x + 50 as evidence of values of a and b.

(b)  3 is a root B1

City of London Academy 14 2 4 200 Solving 3-term quadratic x  or (x  1)2  1  50  0 M1 2

1 7i, 1-7i A1 A1ft4

Note

B1: –3 must be seen in part (b) M1: for solving quadratic following usual conventions A1: for a correct root (simplified as here) and A1ft: for conjugate of first answer. Accept correct answers with no working here. If answers are written down as factors then isw. Must see roots for marks.

Note

ft requires the sum of two non-real conjugate roots and a real root resulting in a real number.

Answers including x are B0 [7]

z 2  8i 1 i 5. (a) 1   M1 z 2 1 i 1 i 2  2i  8i  8  = –3 + 5i A1A13 2

Note

1 i × and attempt to multiply out for M1 1 i –3 for first A1, +5i for second A1

z (b) 1  (3) 2  5 2  34 (or awrt 5.83) M1A1ft2 z 2

Note

Square root required without i for M1 z 1 award M1 for attempt at Pythagoras for both numerator z 2 and denominator

City of London Academy 15 5 5 (c) tan α =  or M1 z1 3 3

arg z 2 = π – 1.03... = 2.11 A12

Note

5 3 tan or tan–1,  or  seen with their 3 and 5 award M1 3 5 2.11 correct answer only award A1 [7]

6. (a) 5 – 2i is a root B11

(b) (x – (5 + 2i)(x – (5 – 2i)) = x2 – 10x + 29 M1M1 x3 – 12x2 + cx + d = (x2 – 10x + 29)(x – 2) M1 c = 49, d = –58 A1, A15

Notes

1st M: Form brackets using (x – α)(x – β) and expand. 2nd M: Achieve a 3-term quadratic with no i’s.

Alternative

Substitute a complex root (usually 5 + 2i) and expand brackets M1 (5 + 2i)3 – 12(5 + 2i)2 + c(5 + 2i) + d = 0 (125 + 150i – 60 – 8i) – 12(25 + 20i – 4) + (5c + 2ci) + d = 0 M1 (2nd M for achieving an expression with no powers of i) Equate real and imaginary parts M1 c = 49, d = –58 A1, A1

(c)

Conjugate pair in 1st and 4th quadrants B1 (symmetrical about real axis) Fully correct, labelled B12 [8]

City of London Academy 16

7. (a)

B11

Note

B1 needs both complex numbers as either points or vectors, in correct quadrants and with ‘reasonably correct’ relative scale

2 2 (b) z1  2  (–1)  5 (or awrt 2.24) M1 A12

Note

M1 Attempt at Pythagoras to find modulus of either complex number

A1 condone correct answer even if negative sign not seen in (–1) term

A0 for ± 5

 1   1  (c)   arctan  or arctan –  M1  2   2 

arg z1 = – 0.46 or 5.82 (awrt) (answer in degrees is A0 unless followed A12 by correct conversion)

Note

3  arctan 2 is M0 unless followed by  arctan 2 or  arctan 2 2 2 Need to be clear that argz = – 0.46 or 5.82 for A1

– 8 9i 2  i (d)  M1 2 – i 2  i

–16 – 8i 18i – 9   – 5 2ii.e a - 5and b  2or – 2 a A1 A1ft3 5 5

City of London Academy 17

Alternative method to part (d)

– 8 + 9i = (2 – i)(a + bi), and so 2a + b = – 8 and 2b – a = 9 M1 and attempt to solve as far as equation in one variable

So a = –5 and b = 2 A1 A1 cao

Note

M1 Multiply numerator and denominator by conjugate of their denominator

A1 for –5 and A1 for 2i (should be simplified)

Alternative scheme for (d) Allow slips in working for first M1 [8]

8. (a) x2 + 4 = 0  x = ki, x = ±2i M1, A1

Solving 3-term quadratic M1

– 8 64 –100 x  = –4 + 3i and –4 – 3i A1 A1ft5 2

Note

Just x = 2i is M1 A0 x = ±2 is M0A0 M1 for solving quadratic follows usual conventions, then A1 for a correct root (simplified as here) and A1ft for conjugate of first answer. Accept correct answers with no working here. Do not give accuracy marks for factors unless followed by roots.

(b) 2i + (–2i) + (–4 + 3i) + (–4 – 3i) = –8 M1 A1cso2

Alternative method : Expands f(x) as quartic and chooses M1 ± coefficient of x3

–8 A1cso

Note

M1 for adding four roots of which at least two are complex conjugates and getting a real answer. A1 for –8 following correct roots or the

City of London Academy 18 alternative method. If any incorrect working in part (a) this A mark will be A0 [7]

l2 – 5i 3– 2i 36 – 24i –15i –10 9. (a) z    M1 2 3 2i 3– 2i l3 = 2 – 3i A12

Note

3 – 2i  for M1 3 – 2i

(b)

P: B1, Q: B1 ft B1, B1ft2

Note

Position of points not clear award B1B0

2 3 (c) grad. OP × grad. OQ =  – 3 2

 = – 1  POQ  ( ) 2

OR POX  tan –1 2 ,QOX  tan –1 2 3 3

2  3 Tan (POQ)  3 2 M1 M1 2 3 1– 3  2

 POQ  ( ) A1 A12 2

Note

City of London Academy 19 Use of calculator / decimals award M1A0

3 2 2(–3) (d) z   i M1 2 2

5 1 = – i A12 2 2

Note

Final answer must be in complex form for A1

2 2  5   1  (e) r      –  M1  2   2 

26  or exact equivalent A12 2

Note

Radius or diameter for M1 [10]

10. (a) 4 B1 1

(b) (x–4)(x2 + 4x + 16) M1A1  4  16  64 x = , x  2  2 3 (or equiv. surd for 2√3) M1, A1 4 2

Root on + ve real axis, one other in correct quad. B1 Third root in conjugate complex position B1ft 2

City of London Academy 20 M1 in part (b) needs (x-“their 4”) times quadratic (x2 + ax +..) or times (x2 +16)

M1 needs solution of three term quadratic

So (x2 + 16) special case, results in B1M1A0M0A0B0B1 possibly

Alternative scheme for (b)

(a + ib)3 = 64, so a3 + 3a2ib + 3a(ib)2 + (ib)3 = 64 and equate real, imaginary parts M1

so a3 – 3ab2 = 64 and 3a2b – b3 = 0 A1

Solve to obtain a = –2, b = 12 M1A1

Alternative ii

(x – 4)(x – a – ib)(x –a + ib) = 0 expand and compare coefficients M1

two of the equations –2a – 4 = 0, 8a + a2 + b2 = 0, 4(a2 + b2) = 64 A1

Solve to obtain a = –2, b = 12

Extra points plotted in part (c) – lose last B mark

Part (c ) answers are independent of part (b) [7]

(a  2i)(a  i) a 2  3ai  2 11. (a) z   M1A1 (a  i)(a  i) a 2 1 a 2  2 1  , 2a 2  4  a 2 1 a  5 (presence of  5 also is A0) M1, A1 4 a 2 1 2

3a  5  (b) Evaluating their “ ”, or “3a”  or 3 5  (ft errors in part a) B1ft 2   a 1  2  3a 3 5 tan  = ( ) , arg z = 1.15 (accept answers which round to 1.15) M1, A1 3 a 2  2 3

B mark is treated here as a method mark

The M1 is for tan (argz) = Imaginary part / real part

answer in degrees is A0

City of London Academy 21 Alternative method:

 1  1 1 (a)   iy(a  i)  a  2i  a  y  a and ay   2 M1A1  2  2 2 1 5 1 5 y  a and ay   a 2   a  5 M1A1 4 2 2 2 2

5 (b) y = (May be seen in part (a)) B1ft 2 tan  5 arg z = 1.15 M1A1 3

Further Alternative method in (b) Use arg(a + 2i) – arg(a – i) B1 = 0.7297 – (–0.4205) = 1.15 M1A1 3 [7]

 2  i 12. (a) (i) Multiply top and bottom by conjugate to give M1A1 5

(ii) Expand and simplify using i2 = –1 to give 3 – 4i M1A1 4

–2 – i or 2 + i OK for method. Attempt to expand required.

(b) z2 – z = 5 – 5i, z 2  z  5 2 * M1A1 2

square root required for method

 (c) arg(z2 – z) =  or –45° or 7/4 or 315° or –0.7853… or 5.497… M1 A1 2 4

2 for correct answer only, tan required for method. 2dp or better.

(d) 2 y 1 x –3–2–1 12 34 5 6 –1 –2 –3 –4 –5 –6 one mark for each point B1, B1 ft 2

Position of points not clear but both quadrants correct first B1 only.

City of London Academy 22 [10]

13. (a) z* = 3 1 B1 z ( 3 i)( 3  i) 3 2 3i 1 1 3   ,   i (*) M1, A1cso3 z * ( 3  i)( 3  i) 31 2 2

M: Multiplying both numerator and denominator by 3 i , and multiplying out brackets with some use of i2 = –1.

2 2 z  1    3   z z 3 1      ,  1 Or :   ,  1 (b)       M1, A12 z *  2   2   z * z * 3 1 

Answer 1 with no working scores both marks.

 iamg( w)   real(w)  (c) arg(w) = arctan   or arg(w)  arctan  ,  real(w)   imag( w)  z where w is z or z* or M1 z *    3  z   2   arg   arctan    A1  z *  1  3  2   1    1   arctan    and arctan   (Ignore interchanged z and z*) A1  3  6  3  6     z  arg z – arg z* =      arg  A14 6 6 3  z *

Allow work in degrees: –60°, –30° and 30° 5 11  Allow arg between 0 and 2: , and (or 300°, 330° and 30°). 3 6 6 Decimals: Allow marks for awrt –1.05 (A1), –0.524 and 0.524 (A1), but then A0 for final mark. (Similarly for 5.24 (A1), 5.76 and 0.524 (A1)).

(d) 2 y

* 1 2 x –1 1 2 3

–1

–2 z and z* (Correct quadrants, approx, symmetrical) B1

City of London Academy 23 z (Strictly inside the triangle shown above) B12 z *

Condone wrong labelling (or lack of labelling), if the intention is clear.

(e) (x ( 3 i))(x ( 3 i)) M1   b   c  Or: Use sum of roots    and product of roots   .  a   a  x 2  2 3x  4 A12 [13]

 2  4  68  2  64i 14. (a) Method for finding z : z  ,  M1, A1 2 2 [Completing the square: (z + 1)2 + 16 = 0, z = 1 16i M1, A1 A13 z = – 1 ± 4i (a = – 1, b = ± 4)

First A1 is unsimplified but requires i –1 ± 8i only scores M1 unless intermediate step seen when M1A1 possible Correct answer with no working is full marks

SC: If M0 awarded, k ± 4i, k + 4i , k – 4i scores B1 (Epen M0A0A1)

Use of z = a + i b

(i) z2 – 2az + a2 + b2 = z2 + 2z + 17 = 0 and compare coefficients M1 a2 + b2 = 17 and a = –1; z = – 1 ± 4i A1, A1

(ii) (a + ib)2 + 2(a + ib) + 17 = 0 and compare coefficients M1 2b(a + 1) = 0 and a2 – b2 + 2a = – 17, a = –1 and b = ±4 A1, A1

(b) y 4

x –1

–4 B1ft1

Must be a conjugate pair.

Allow: Coords marked at points or “correct” numbers on axes. (allow “graduations”) (Ignore any lines drawn) [4]

City of London Academy 24

z 1 pi (5  3i) 15. (a) 2  . M1 z1 5  3i (5  3i) 5  5pi  3i  3p  [Multiply out and attempt use of i2 = – 1] M1 (34) 5  3p 5p  3 5  3p 3 3p   i or  i A13 34 34 34 34

z If 1 used treat as MR. Can score (a)M1M1A0 (b)M1A0 z2  5  3p 3 5p 1 (a)  i (b)   2 2   1 p 1 p 4 Allow A1 if answer “all over” 34, real and imag. collected up) 1 + pi = (a + ib) (5 + 3i): M1 compare real and imag. is first M mark If denominator in (a) incorrect, both marks in (b) still available

z d  (b) For 2 = c + id using  tan : M1 z1 c 4 [5p – 3 = 5 + 3p]  p = 4 A12

 In (b), if use arg z2 – arg z1 = : 4

3   M1 for arctanp – arctan  [arctan p = + 0.5404... = 1.3258] 5 4 4

Allow A1 for p = 4 without further work or for that shown in brackets, i.e. assume values retained on calculator (no penalty because it looks as though not exact) [5]

16. (a) 2z + iw = l iz  iw = 3i  3

Adding 2z + iz = 4 + 3i Eliminating either variable M1

 4  3i z A1 2  i

 4  3i 2 i z   M1 2  i 2 i

 8 3 4i  6i  5

City of London Academy 25 = – 1 + 2i A1 4

(b) arg z =   arctan 2 M1

 2.03 cao A1 2 [6]

17. (a) 3 + 2i is a solution B1

(x  3  2i)(x  3 + 2i) = x2  6x + 13 M1

f (x) = (x2  6x +13)(x2 + ax + b)

b = 6 B1

Coefficients of x3 a  6 = 6 or equivalent M1 a = 0 A1

x2 + 6 = 0  x = 6i, 6i M1A1 7

(b)

× ×  o

× × 

Conjugate complex pair on imaginary axis B1

Conjugate complex pair in correct quadrants B1 2 [9]

  2i 18. (a) z + 2i = iz +  (1 – i)z =  – 2i, z = M1, A1 1 - i   2i 1 i 1 z =  ,  (...... ) M1, A1 1 i 1 i 2   =   1   1i A1 cso 5  2   2 

City of London Academy 26  1  1 nd 2 (b) 2  1 ,   6 2 M: Solving = k (constant k) M1 M1 A1 3  1 2   1 2 2

(c) z = + 2i, |z|2 = 42 +22 = 20 M: Subs. λ value and attempt |z| or |z|2M1 A1 2 [10] Alternative (a): u + iv + 2i = iu – v +  Real parts: u =  – v M1 A1 Imaginary parts: v + 2 = u A1  v  1 Solve for u or v M1 2  v  1 Both correct A1 2

Alternative (a): (working back)              1  i 1  1  i 1  1  i 1  2   2   2   2   2   2  M1  M1                  1  i 1  1  i 1  1  i 1  2   2   2   2   2   2 

   2 2 Numerator: 2i  1 A1, Denominator: 2 A1, = i A1 cso  2 

(b) ‘Upside down’ (should give λ = –6) scores M0 M1 A0, and can continue to score M1 A1 in (c).

19. (a) 1 – 3i is a root B1 (z – 1 – 3i)(z – 1 + 3i)(z + ) = (z2 – 2z + 10)(z + ) = z3 + 6z + 20 10 = 20   = 2  –2 is a root M1 A1 3 Mi any complete method

B (b) A

C Position of points only in correct quadrants and negatives x-axis B1ft 1

3 (c) mAB = = 1, mAC = –1 M1 3

City of London Academy 27 Full method

mAB.mAC = – 1  triangle is right-angled A1 2 [6]

Alternatives

AB2 + AC2 = 18 + 18 = 36 = BC2 Result follows by (converse of ) Pythagoras, or any complete method M1 A1

 4 6 20. (a)  arctan or  – arctan or equiv. in degrees M1 2 6 4 arg z = 2.159 A1 c.a.o. 2

A (b) w = 20  20  M1 5 Full method for A using w = 20

 A = 10 A1 A  2  i  w =    , w = 4 + 2i M1, A1 4 2  i  2  i 

 w   2  (c) arg   = arg w – arg z = arctan   – (a) M1  z   4  = 0.463… – 2.159 A1ft(a) 2dp or better

= – 1.695… A1 3 awrt –1.70 [9]

Alternative (c) w = –0.0769… –0.6153….i M1 z Attempt w ÷ z and use arctan

 w   0.6153...  arg     arctan  A1  z    0.0769... Expression (2dp)

= awrt –1.70 A1 3

City of London Academy 28 21. (a) Second root = 3 – i B1

Finding product of two roots (= 10), or quadratic factor (x2 – 6x + 10) M1

Complete method for third root or linear factor M1

Third root = ½ A1 4

(b) Using conadidate’s three roots to find cubic with real coefficients M1 (= 2x3 – 13x2 + 26x – 10)]

Equating coefficients M1 a = – 13, b = 26 A1 3 [7]

z z 4 22. (a)  ;  = 2 M1M1A1 3 w w 2 [M1 for correct modulus, M1 division of moduli]

 z  (b) arg   = arg z – arg w M1  w  3    13  11 =      ;  M1A1 3 4  3   12  12 [Second M1 for one correct arg]

…………………………………………………………………………………………………………....

 z  Working with   :  w 

 z  2 2(1  i) 2 2(1  i)(1  i 3) (a)     M1  w  1  i 3 4  2{(1 3)  i( 3 1)}    2  Correct method for finding modulus, = 2 M1A1 3

"( 3 1)" (b) Finding tan 1 M1 "( 3 1)"

City of London Academy 29  z  Complete method for arg   M1  w  11 =  A1 3 12

…………………………………………………………………………………………………………....

(D) C

For A B1 For B B1 For C B1ft 3

 (d) DOB = or 60° B1 3 Correct method for AOC M1    AOC =   A1 3 4 12 3 (cso)

1  (e) Area AOC = × “4” × “2” × sin“ ” = 2 3 (3.46 or better) M1A1 2 2 3 [14]

23. (a) z = a + ib  (a2 – b2) + 2abi = – 16 + 30i M1 Equating imaginary parts 2ab = 30 and thus ab = 15 (*) A1 2

(b) Also (a2 – b2) = – 16 B1 Attempt to solve by valid method involving elimination of unknown M1 z = 3 + 5i or z = – 3 – 5i A1 A1 4 [6]

City of London Academy 30

24. (a) w = (1 + 3 i)(2 + 2i) M1 = (2 – 2 ), (2 + 2)I A1, A1 3

   2 3  2  (b) arg w =arctan   or adds two args e.g. 60° + 45° M1  2  2 3  7 = or 105° or 1.83 radians A1 2 12

Im B

(d) Re B1B1 2

f.t. w in quadrant other than first

(e) AB2 = 4 + 32 – 16 cos45 (=20), then square root M1 AB = 2 5 A1 2 Or w – z = 1 – 2 + i(2 + )  AB = w – z = (1 2 3) 2  (2  3) 2 M1 = 20 = 2 A1c.a.o. 2 [13]

25. (a) | z | = 22 | w | = 2; M1, A1 | wz2 | = (22)2 × 2 = 16 M1, A1  5 2  5  Arg z = – 4 Arg w = 6 ;  Arg wz = – 4 – + 6 = 3 , 60° M1, A1 6

OR z2 = –8i; z2w = 8 + 83i M1, A1 | z2w | = 82  82  3 M1

City of London Academy 31 = 16 A1 arg z2w = tan–13 M1  = 3 A1

× C

(b) B ×

× A Points A and B B1 Point C B1ft 5  BOˆC =  M1 6 3  = or 90° A1 4 2 [10]

26. (a) x3 – 27 = (x – 3)(x2 + 3x + 9) M1 (x = 3 is one root). Two roots satisfy x2 + 3x + 9 = 0 (*) A1 2

(b) x = 3 B1  3  9  36 x = M1 2 3 3 3 =   i A1 3 2 2  3  3 3i accept 2

y × (c) 0 x ×

root on positive real axis and one other root in correct quadrant. B1 third root in conjugate complex position. B1 ft 2 [7]

City of London Academy 32

4  3i (4  3i)(2  4i) 20 10i 27. (a) = = (= 1 – 1 i) M1 2  4i (2  4i)(2 – 4i) 20 2 5 |z| = 12  ( 1 )2 , = M1, A1 3 2 2 awrt 1.12, accept exact equivalent

5a  (6  a 2 )i (a  3i)(2  ai) (b) = 4  a 2 M1 (2  ai)(2  ai) accept in (a) if clearly applied to (b)

6  a2 (tan  ) 1 = M1 A1 4 5a leading to a2 + 5a – 6 = (a + 6)(a – 1) = 0 a = –6, 1 M1 A1 a = –6 is in wrong quadrant; reject –6, a = 1 is only answer A1 6 [9]

Alternatives

(a) |4 + 3i| = 5, |2 + 4i| = 20 M1 5 |z| = (= , or exact equivalent) M1 A1 3 20

(b) arg z = arg (a + 3i) – arg (2 + ai)  3 a  arctan  arctan M1 4 a 2 3 a  1 = a 2 3 1 2 then as before M1 A1 3 [6]

28. zw =

  2  2    2  2  12 cos cos  sin sin  + 12i sin cos  cos sin  B1 for 12  4 3 4 3   4 3 4 3 

 11 11  = 12 cos  i sin  M1 A1  12 12 

City of London Academy 33 [3]

29. (a) z2 = (3 – 3i)(3 – 3i) = 18i M1 A1 2

1 (3  3i) 3  3i 1 i (b) = = = M1 A1 2 z (3  3i)(3  3i) 18 6

1 1 1 2 = = = z 18 3 2 6 all three correct A1 2

C × D × O A ×

(d)

B×

two correct B1 four correct B1 2

OB OA 3 2 (e) = 18,  = 18 M1 A1 OD OC 2 / 6 AOB = COD = 45  similar B1 3 [11]

30. (a) Clearly by inspection z – 1 is a factor of f(z) B1

z3 – 1 = (z – 1)(z2 + z + 1) B1

1 (1 4) z2 + z + 1 = 0  z = M1 2

City of London Academy 34 1 i 3 1 i 3 so z =   or   A1 (both)4 2 2 2 2

(b)

1 3 – + i 2 2 I

R O 1

1 3  – i G11 2 2

2 2  1 3   1   1  3   3    i    2 (c)   =    2 i    i M1  2 2   2   2  2   4 

1 3 3 1 3 =  i  =  i = other root A1 4 2 4 2 2

2    1 3  1 3 Similarly,   i  =   i A13  2 2  2 2 [8]

1. No Report available for this question.

2. No Report available for this question.

3. Most candidates found this question very accessible with many scoring 7 marks. In part (a) a minority of candidates 2 failed to appreciate this was a “show that” question and omitted the key step of stating or using i = –1. Other than this, numerical errors were very rare. Part (b) was very well done and the modulus was usually correctly given as 13. In part (c) most candidates appreciated that inverse tangent was needed, but many could not deal with the fact that the point was in the third quadrant. Wrong answers such as 1.97 and 1.18 were common. The need to identify the correct quadrant is essential for candidates hoping to continue to FP2 and FP3. In addition, a common error was to give the answer as -1.96, arising from rounding too early when finding – + 1.18. In part (d) the Argand diagram was usually correct, though there were some errors. Candidates should be advised not to extend their working to the very bottom of the page, past the scanned area. Many plots of –5 – 12i were beyond the scanned area.

4. In part (a) long division and comparing coefficients were each used to good effect and errors were rare. Part (b) resulted in many good answers. Some however felt that 3, and even x + 3, was a root and others omitted the real root completely.

City of London Academy 35 Some confused roots with factors. It was disappointing at this level to see many candidates failing to solve a quadratic correctly. Candidates should be advised to quote the quadratic formula before using it to ensure that they earn the method mark. Completion of the square was often more successful in this question than use of the formula. In part (c) some included an x in their answer and others found a product instead of a sum. The vast majority earned this follow through mark however.

5. This question was well answered with many candidates gaining full marks. In part (a) there were very few candidates who were unaware of the technique of multiplying numerator and denominator by (1 + i). Careless mistakes with arithmetic were the main reason that candidates lost marks on this part of the question e.g. dividing by 2 to give –6 + 5i or –3 + 10i. These were rare however. Part (b) was also well answered with most candidates following the method outlined in the mark scheme although some candidates correctly answered 2 2 the question by calculating mod z1 divided by mod z2. Unfortunately √(–3 + 5i ) was occasionally seen. Candidates who were successful in part (c) employed a variety of methods including π/2 + arctan(3/5), π – arctan(5/3), π + arctan(5/-3) and arg z1 – arg z2

6. This question was generally very well done. Nearly every candidate achieved the complex conjugate in part (a). In part (b) most candidates chose the method of expanding brackets, then equating coefficients to achieve c and d. This was generally very well attempted. Most mistakes were algebraic ones, with methods clearly understood. More confusion reigned amongst those who chose the alternative method. Some substituted 2 and found themselves with one equation in 2 unknowns. Those who substituted a complex root often made mistakes with powers of i or did not use the concept of equating real and imaginary parts. The Argand diagrams in part (c) were good, and most found some way of introducing scale to their diagram, usually by “vectors”, co-ordinates, or labelling the axes.

7. Almost all candidates achieved the mark in part (a) for the argand diagram. Also the modulus of a complex number was understood and usually found correctly. Finding the argument of the complex number caused more problems for some candidates as a number of them did not consider which quadrant they needed. Also some candidates used incorrect trigonometry. A few answers were given in degrees and some calculated tan (0.5) instead of arctan (0.5). Part (d), which involved calculating a quotient, was usually answered correctly also. Most successfully multiplied by the conjugate 2 – z1 i and got full marks. Some misread this part and found which meant that they could only get a maximum of 1 z2 mark for multiplying by their conjugate.

8. A substantial minority multiplied out the two brackets which complicated the problem. Most however attempted to 2 solve x + 4 = 0, but there were a number of wrong answers, particularly the real answers +2 and –2. The solution of the three term quadratic was usually correct but there were errors in simplification with a substantial number of candidates losing some accuracy in part (a). Indeed the answers –4 + 6i and –4 – 6i were fairly common. In part (b) candidates were asked to find the sum of their roots. Most obtained –8, but this only gained M1A1 if it followed wrong roots. There were candidates who were unfamiliar with the term “sum” and found the product instead.

9. In part (a) the majority of candidates knew about multiplying by the conjugate and very few failed to complete the arithmetic accurately to obtain 2 – 3i. In part (b) the quality of the diagrams varied a great deal. Some were carefully done using rulers and were easy to read. Others were little more than rough sketches, although using coordinates or a scale meant that the positions of both P and Q were usually clear. A variety of methods were seen in part (c). The most popular was to use arctan and decimals. A few tried the alternative of gradients and were successful as were those who attempted the converse of Pythagoras’ theorem. A minority of candidates took a geometric approach based on similar triangles with mixed success. A significant number of candidates did not seem to know the ‘angle in a semicircle’ result in part (d) to enable them to deduce that PQ was a diameter which made finding the centre and radius relatively easy. Some let C be (a,b) and formed two quadratic simultaneous equations by equating OC, PC and QC. Not all were able to complete accurately. Another method was to find the equations of the perpendicular bisectors of OP and OQ and find their point of intersection. Again some were successful but many floundered in the algebra. A significant number of candidates did not attempt part (d) or part (e).

10. In part (a) almost every-one earned the mark for the answer 4, but incorrect answers included 2, 8, and –4. 3 2 The most common approach to part (b) was to factorise (x – 64) into (x – 4)(x + 4x + 16), and then to solve the 2 quadratic equation (x + 4x + 16) = 0. Most who did this were successful and obtained the answers  2  2 3 . Mistakes included slips in long division with some attempts failing to reach the quadratic factor, and the solution of the quadratic equation (when obtained) caused further errors for some trying to complete the square. Other methods of 3 solution began with (a + ib) = 64, expanded with a binomial expansion and equated real and imaginary parts. Another method was to expand (x – 4)(x – a – ib)(x – a – ib) = 0 and to compare with 3 x – 64 = 0. The latter two methods seemed to give rise to more errors than the first method described.

Part (c) usually resulted in full marks, as the mark scheme was generous here and did not require correct labelling on the axes. The second of the two marks was a follow through mark for having the third root in a conjugate complex position to the second. A few candidates strangely drew their real axis vertically and their imaginary axis horizontally.

City of London Academy 36 Candidates should be discouraged from this practice. As usual, the complex numbers were represented by points, line segments or vectors on the diagram.

11. This question discriminated between those who understood complex numbers and those who just followed the algorithms that they had been taught. Most understood that they were required to express the quotient as a single complex number by multiplying the numerator and the denominator by the conjugate of the denominator. They then needed to find the real part and to put it equal to a half. However a minority did not realise that this stage was required and merely put the complex expression that had been given equal to a half, thus showing little understanding of the 2 question and gaining no credit. Common errors were a – 1 on the denominator, and algebraic slips leading to a = √3. Some lost the final mark in (a) by including the negative root. The alternative solution given on the mark scheme was rarely seen.

Part (b) was usually done well. Most found their imaginary part and the “follow through” enable most to gain this mark. imaginary part Some candidates did not make their working clear for finding the argument and a few used tan 1 ( ) 5 confusing their value of a with the real part of the complex number. A 5 small minority found the modulus rather than the argument, and some used tan rather than arctan. Almost all gave their answer in radians to 2dp. Again, very few used the alternative method on the Mark Scheme.

12. This was a very accessible question for the majority of candidates with many scoring full marks. Occasionally in part (a) candidates thought 1/(–2 + i) was equal to –1/2 + 1/i but these errors were very infrequent. Almost all were able to 2 expand successfully and use i = –1 to simplify their answer. In part (b) some subtracted incorrectly but gained a method mark for knowing how to find the modulus of a complex number. The most common errors were in part (c) where the argument was given as π/4 or 3π/4 instead of –π/4. Also in part (c) some tried to use arg(z/w) = argz – argw. Marks were lost in part (d) for poorly labelled or ambiguous Argand diagrams.

13. The independence of different parts made it possible for most candidates to pick up marks throughout this question for their knowledge of various different aspects of complex number theory. In part (a), just a few candidates did not understand ‘complex conjugate’, but the vast majority were able to obtain the given result without very much difficulty.

2 2 z  1   3  z z    The value of in part (b) was usually found from     , but sometimes by using  . The z *  2   2  z * z *

2 2  1   i 3     mistake     was occasionally seen.  2   2  Part (c) proved difficult for the average candidate. For some, there was much confusion in the definition of ‘argument’, with 1 1 statements such as arg z =  or arg z = tan , but ‘recovery’ was sometimes possible, perhaps with the help 3 3 2 5 of a diagram. Common mistakes included the omission of minus signs and the use of and as arguments 3 6   instead of  and  . 3 6 z In the Argand diagram in part (d), z and z* were usually correctly shown, but the position of was sometimes wrong. z * Attempts at part (e) were often very good, the usual method being to expand x   3 ix   3  i. The use of ‘sum and product of roots’, a very efficient method here, was comparatively rare.

14. Virtually all candidates completed this question and, although there were some common mistakes, notably simplifying  2   64  2  8i , or even , to –1 ± 8i, this proved a good starter for most candidates. 2 2 The most common approach was to use the quadratic formula, although the given equation was very easily and successfully solved by “completing the square”. Other routes taken were to substitute z = a + ib into the given equation or compare {z – (a + ib)}{z – (a – ib)}= 0 with the given equation and, although these were more time consuming solutions, it is good to report that these were usually very competently handled.

15. Another question where the majority of candidates were able to score well, particularly in part (a). Errors here were generally slips such as (pi)(–3i) becoming 3, not 3p, and the one that all examiners reported on, 36 (from 25 + 9) in the denominator being surprisingly common.

City of London Academy 37 b  In part (b) candidates who used  tan , were generally more successful than those who used arg z2 – arg z1 = a 4  θ . In the latter case there was confusion between whether arg z = tan or 4 –1 3  tan θ, with statements like tan p – tan  = , and poor manipulation, such as 5 4 –1 –1  3   3 tan p – tan     p   1 , common mistakes.  5  4 5

16. This question proved a very accessible start to the paper and almost all candidates could use an appropriate method to solve the pair of simultaneous equations. The majority used methods of elimination or substitution they had learnt for GCSE. Completely correct solutions to part (a) were common but, as noted above, many spoilt their work with inaccurate algebra. It was quite usual to see two or more mistakes in elementary algebra, particularly in signs. A significant minority used the alternative method of substituting, say, z = a ib and w = c id and, by equating real and imaginary parts, obtained 4 equations in 4 unknowns. Superficially this seemed a complicated method but, in practice, the equations came out quite easily and completely correct solutions using this method were not uncommon. In part (b), the majority could use a tangent to find an angle related to the argument but getting this angle into radians in the right quadrant proved demanding. It is very helpful to draw a diagram in such circumstances and candidates should be encouraged to do this.

 2   64  2  8i annotation reference; This 2 2 17. proved to be the easiest question on the paper. A few had difficulty in finding the correct second quadratic factor but full 2 marks were common. An unexpected source of error was that a number who had the correct second factor x +6 were 2 unable to solve x + 6 = 0 correctly. x = ±√6 and x = ±6i were both relatively common. Almost all could show their answers on an Argand diagram correctly. a – ib)}= 0 with the given equation and, although these were more time consuming solutions, it is good to report that these were usually very competently handled.

18. There were many excellent solutions to this question, but where candidates did find difficulty it was usually in part (a). z  2i z  i Attempts that started with  , effectively treating z as a real number, were doomed to failure and often z i z  i led to much confusion and wasted time. Fortunately, however, candidates could complete parts (b) and (c) independently of (a), and usually did so successfully.

 1  2 1 2 1 1 Occasional mistakes in (b) included  and  arctan . .  1 2  1 2 2 2 Also in part (b), there were a few candidates who equated the imaginary part to 1 and the real part to 2, giving two different values for λ.

19. Part (a) was usually answered well. The root 1 – 3i was identified as part of a conjugate pair and then a variety of rd methods were used to find the 3 root. Forming a quadratic factor and using long division was a popular approach but a number used the factor theorem and the more astute spotted that the sum of the roots was equal to zero thus enabling them to write down the answer very quickly. Some candidates seemed unsure of the difference between a factor and a root. The sketch in part (b) was usually completed successfully, but when the three points clearly did not form a right angled triangle (as when z = 2 was given as the third root in part (a)) this did not seem to prompt candidates to check their earlier working. A variety of methods were employed to answer part (c) with gradients, the inverse of Pythagoras’ theorem and the identification of two right angled isosceles triangles being the most common. A few candidates used the scalar product with a vector approach which worked well. Some candidates lost the final mark in part (c) because their proof was incomplete: assuming, rather than stating and justifying, symmetry was a common fault.

20. Most candidates knew that the argument of a complex number had something to do with the arctan function but few drew a diagram or knew how to obtain a value in the correct quadrant and this affected their answers to parts (a) and (c). Part (a) was not answered well with few candidates having a correct strategy for obtaining an argument in the second quadrant. There were some alarming algebraic errors seen in part (b) and often these were not connected with the complex numbers but evidence of a poor grasp of the basic rules that should have been securely established at this A AA stage; for example when became  . However a large number of candidates did multiply w by 2i 2i

City of London Academy 38 (2 i) and often they knew how to find the modulus of a complex number and could use w  20 to establish (2 i) that A = 10 and hence find w. Some candidates multiplied w by (2 – i) and then by setting the imaginary part to 0 were able to find w without needing to evaluate A. w In part (c) most used arg arg(wz ) arg( ) and were able to make some progress, but a number preferred z w the longer method of finding in the form a + ib and then trying to find the argument of this complex number. z Unfortunately these candidates often ignored any consideration of an appropriate quadrant for their argument.

21. The knowledge of complex roots appearing in conjugate pairs was almost universally known. There were many approaches to this question: some candidates tackled part (a) first, others chose part (b) first, and many combined the two. Those candidates who tackled part (a) first usually went on to correctly find the quadratic factor x2  6x  10 . The third root was then an easy step, although the linear factor 2x – 1 was frequently left as the root. Many candidates, however, at this stage could not resist long division. The difference between 3 2 2 2x + ax + bx – 10  (x – 6x + 10)(2x – 1) and 2x + (a + 12) 2 3 2 x – 6x + 10 2x + ax + bx –10 3 2 2x – 12x + 20x 2 (a + 12)x + (b – 20)x – 10 2 (a 12)x – 6(a + 12)x + 10(a + 12) (52 + b + 6a)x – (10a + 130) is clear to see; not only is the latter very time-consuming it is also open to more errors. Of course, a and b can also be found from this work, but few candidates taking this route were completely successful. Even when a had been found candidates often forgot to go back to answer part (a) and state the third root. Another common approach was to tackle part (b) first by setting f(3 + i) = 0 and solving the resulting simultaneous equations in a and b. Correct solutions were 3 seen but again errors were common, particularly in simplifying (3 + i) .

22. For candidates who had time to consider this question seriously marks were readily available, although methods were z often long-winded, particularly in expressing in the a + ib form, when both parts (a) and (b) became more testing. w z z  z  Those candidates who used  and arg    arg z  arg w , were able to give succinct answers, but w w  w  even here there was not a widespread appreciation that the form of z displayed both z and arg z , and arg z was often 13 11 given as  rather than –  . In part (c) a large number of candidates did not realise the significance of their 12 12

answers to parts (a) and (b) and proceeded to find in the form a + ib before they could plot C. Many candidates

were able to gain some marks in part (d) although some thought that demonstrating that two pairs of sides of equal length in triangles AOC and DOB was sufficient to prove congruency and hence the equal angles. Part (e) provided two easy marks for many candidates. Those candidates who assumed that triangles AOC and DOB were right angled did not gain full marks in parts (d) and (e) unless they justified the fact.

23. The response to this question was disappointing. Most candidates were able to expand (a+ib)(a-ib) accurately and to simplify (ib)2 as –b2 not +b2, (a few gave this as just b or –b), and to compare imaginary parts correctly. There was much carelessness in (b), from getting a2-b2=+16, to eliminating a or b, to getting the wrong signs in their factorisation of the resulting quadratic in a2 or b2 e.g. a4+16a2 – 225 = (a2-25)(a2+9). When using the formula to solve their equation, many forgot that they were solving for a2, or b2, and did not go on to obtain the correct, appropriate values of a and b. Many also disregarded the information that a, b were real numbers, and gave 4 solutions to the given quadratic equation. There was a common assumption that a first root of 3+5i implies that the second root was the complex conjugate 3 - 5i, as the fact that the given equation did not have real coefficients was overlooked.

24. On the whole, (a) was correctly done and was done by the expected method. Quite a few however went for the option of dividing a + ib by 1 + √3i. This involved more work, but was usually successful. In (b), the appropriate quadrant for the solution to arg w was often totally ignored, although the correct placing of B in the second quadrant then appeared in (d). Very few candidates added two arguments to obtain their answer. Other common errors were to take tan(correct ratio), rather than arctan … , or to get in a muddle working with π/2. Some candidates generated a wrong answer by

City of London Academy 39 trying to rationalise the denominator of their fraction involving surds before looking for arctan(their value). Part (c) was usually correct but very few multiplied moduli to give the answer. A few candidates confused modulus and argument. In (e) there was much carelessness in the calculation of the difference in coordinates of A and B, particularly in the x coordinates . These were often given as (1 + 2 - 2√3), or the results from (a) were interchanged so A became (2 + 2√3, 2 - √3). Candidates did not always observe the instruction to give their answer as a simplified surd, either stopping with √20 or giving a decimal answer.

25. This was a popular question with many candidates gaining full marks. In part (b) many candidates drew an accurate diagram on graph paper when a sketch was sufficient.

26. Part (a) proved the most difficult part of this question. Even those who did succeed in factorising often did not state what conclusion this led to. However the majority could either use the information in part (a) or an independent method 33   to solve parts (b) and (c). Even if slips were made, and the error   3i  was unexpectedly  2 common, any reasonable results were followed through in part (c).

27. Part (a) was almost invariably correct but part (b) proved demanding. Many attempts were spoilt by inaccurate algebra associated with multiplying by the conjugate complex 2 3i . Those who used the conjugate complex correctly 6  a2 6  a2  could usually reach an equation of the form  1 , although  was not infrequently seen. With 5a 54a the correct equation the solutions a = 1 and a = 6 usually followed but many were unable to give a sound reason for rejecting 6. Those who attempted the alternative method using argz arg( a  3i)  arg(2  a i) were rarely successful. This method, although, of course, mathematically sound, leads to the equation  3 a arctan arctan 42a and solving this proved beyond all but the strongest candidates.

28. Many candidates were familiar with the rules for modulus and argument of a product and simply wrote down the answer. Those who attempted to use surds rarely managed to change back to the modulus-argument form successfully.

29. This was a popular question with most candidates able to attempt the first four parts. Several candidates had difficulty in establishing the similar triangles. Many lost accuracy by using decimals for angles and ratios. A majority of candidates thought that it was sufficient to find a pair of corresponding sides with a common ratio to prove similarity.

30. No Report available for this question.

City of London Academy 40