MA1006 ALGEBRA – List 5 - SOLUTIONS

1. Read the wikipedia articles on Euler’s formula, and roots of unity.

2. On the draw the circle of all complex of modulus one and on this circle mark all roots of unity of degree n 2,3,4,5,6,7,8,9,10 (make nine separate drawings). = On each figure pick two roots of unity and multiply them. Pick a root and observe the behaviour of its powers. Repeat experiments until you observe different patterns depending on whether or not the degree n is prime.

3. Let z,w C be two roots of unity of fixed degree n N. Prove that their product is also a root ∈ 1 ∈ of unity. Prove that the inverse z− is a degree n root of unity. In the polar form, the roots of degree n of unity are the complex numbers µ ¶ i(k 2π ) ³ 2π´ ³ 2π´ wk e n cos k i sin k , where k 0,...,n 1. = = n + n = −

Choose any two of these roots, wk and w j . Then, by De Moivre’s Theorem,

i(k 2π ) i(j 2π ) i((k j) 2π ) w w e n + n e + n . k j = = i((k j) 2π ) If we denote this product by w e + n , then, = n i(n(k j) 2π ) i((k j)2π) w e + n e + 1. = = = In other words, w is also a root of degree n of unity. Notice that ½ wk j , if k j n 1 w + + ≤ − = wk j n, if k j n + − + ≥

1 Fix now a root of degree n of unity, w . If you choose w 1, then obviously w − 1, which k 0 = 0 = is again a root of unity. Thus, suppose 0 k n 1. The inverse of w is a complex < ≤ − k z z eiα such that = | | i0 i(α k 2π ) e 1 zw z e + n = = k = | | Thus, z 1 and α 2π k 2π (n k) 2π . Since 0 k n 1, 0 n k n 1. Thus, z is also | | = = − n = − n < ≤ − < − ≤ − a root of unity of degree n: z wn k . = − 4. Let n N be a natural number. Suppose that a natural number k divides n. Show that the ∈ set of all degree n roots of unity contains all degree k roots of unity. Draw relevant figures for small values of n, say, up to 12.

If k divides n, then n kq for some q N. Write again the roots of unity of degree n as = ∈ i(j 2π ) w e n , where j 0,...,n 1. j = = − Then, w0 1 is obviously a root of unity of degree k. Let’s see what happens now with i(q 2π=) w e n . Using De Moivre’s Theorem, q = k i(kq 2π ) i(2π) w e n e 1, q = = =

1 because n kq. That is, w is a root of unity of degree k. You can easily check that the = q same is true for w2q ,...,w(k 1)q . − This means that the set of roots of unity of degree n contains k different roots of unity of degree k. On the other hand, you know that there are only k different roots of unity of degree n. Thus, the exercise is finished.

5. Give an example of a of modulus one that is not a root of unity.

A root of unity is a complex number z such that zn 1 for some n N. Then, for instance = ∈ z eip2 is not a root of unity because there is no natural number n such that zn ei(np2) 1 = = = because np2 is never a multiple of 2π.

6. List all degree two and three roots of the following complex numbers (use polar coordinates):

a. 1 ei0; = degree 2 degree 3

w 1 z 1 0 = 0 = w ei(π) 1 z ei(2π/3) 1 p3 i 1 = = − 1 = = − 2 + 2 z ei(4π/3) 1 p3 i 2 = = − 2 − 2 b. i ei(π/2); = degree 2 degree 3

w ei(π/4) p2 p2 i z ei(π/6) p3 1 i 0 = = 2 + 2 0 = = 2 + 2 w ei(5π/4) p2 p2 i z ei(5π/6) p3 1 i 1 = = − 2 − 2 1 = = − 2 + 2 z ei(9π/6) ei(3π/2) i 2 = = = −

c. 1 i 2ei(π/4); + = degree 2 degree 3

w p2ei(π/8) z p3 2ei(π/12) 0 = 0 = w p2ei(9π/8) z p3 2ei(3π/4) 1 = 1 = z p3 2ei(17π/12) 2 =

d. 2 i p5eiα, where α arctan(1/2); + = = degree 2 degree 3

w p4 5ei(α/2) z p6 5ei(α/3) 0 = 0 = 4 i(α/2 π) 6 i(α/3 2π/3) w p5e + z p5e + 1 = 1 = 6 i(α/3 4π/3) z p5e + 2 =

2 7. Find all solutions of the x4 x3 x2 x 1 0. Consider first the equation + + + + = x5 1 0 − = Clearly, x 1 is a solution for this equation, and then we know that x 1 divides x5 1: = − − x5 1 (x 1)(x4 x3 x2 x 1). − = − + + + + It follows now that the solutions of x4 x3 x2 x 1 0 are the other roots of unity of degree + + + + = 5: w ei(2π/5) w ei(4π/5) 1 = 2 = w ei(6π/5) w ei(8π/5) 3 = 4 = 8. Prove that the sum of all degree n roots of unity is equal to zero. Obviously, we have to assume that n 2. As usual write the roots of unity of degree n as ≥ µ ¶ i(k 2π ) ³ 2π´ ³ 2π´ wk e n cos k i sin k , where k 0,...,n 1. = = n + n = −

Now, since n 2, we have w 1, and then, for each k, ≥ 1 6= ½ wk 1, if 0 k n 2 w w + ≤ ≤ − 1 k = w , if k 0 0 =

Pn 1 Let A k−0 wk . Then, = = µ n 1 ¶ X− w1A w1 wk w1w0 w1w1 ...w1wn 1 = k 0 = + + − = = n 1 X− w1 w2 ... wn 1 w0 wk A. = + + + − + = k 0 = = Thus, A satisfies A(1 w ) 0. Since w 1, it follows that A 0. − 1 = 1 6= = 9. Let S {z C : z 1 and 0 arg(z) π/4} be a circular sector. Draw it on the complex plane. = ∈ | | ≤ ≤ ≤ Describe and draw the image of S with respect to a function f : C C defined by: → a. f (z) 2z; since the argument of 2 as a complex number is zero, we have, for each z C, = ∈ 2z 2 z eiα, = | | where α is the argument of z. Thus, the function f (z) 2z is a scaling of factor 2. = b. f (z) i z; in this case, i eiπ/2. Thus, for each z C, = = ∈ iπ/2 iα i(α π/2) i z z e e z e + , = | | = | | and this is a rotation of angle π/2.

c. f (z) z; this is the same as multiplying by 1 eiπ. Thus, this function is a rotation = − − = of angle π.

3 d. f (z) 2i z; this function is the composition of the functions g(z) i z and h(z) 2z. = = = Thus, f (z) 2i z is a rotation of angle π/2 followed by a scaling of factor 2. = e. f (z) (i 1)z; if we write i 1 in polar form, we have i 1 p2ei(π/4). Thus, for each = + + + = z C, ∈ i(π/4) iα i(π/4) iα i(α π/4) (i 1)z p2e z e p2 z e e p2 z e + . + = | | = | | = | | This is a rotation of angle π/4 followed by a scaling of factor p2.

f. f (z) z i; if we write z in Cartesian form, we have z a bi and then = + = + z i a bi i a (b 1)i. + = + + = + + Thus, this function is a translation in the direction of the vertical axis.

g. f (z) z2; write z in the polar form: z z eiα. Then, = = | | z2 z 2ei(2α). = | | This function rotates each complex number by its own argument, and then applies a scaling of factor z . Notice that this function sends the set | | C {z C z 1} = ∈ | | | = to the set C itself.

h. f (z) z3; this is similar to the previous one: = z3 z 3ei(3α). = | | Again, this function sends the set C above to itself.

i. f (z) zn for all natural numbers n N; similar to the previous two parts. = ∈ 1 j. f (z) z− ; notice first that this function is not defined for z 0. For z 0, we have z iα= 1 1 i(2π α) = 6= = z e and z− z − e − . This function is a reflection with respect to the horizontal | | = | | 2 axis, followed by a scaling of factor z − (the exponent 2 is not a mistake, just in case). | | − Notice that, again, the function f sends the set C of complex numbers of modulus 1 to the set C itself.

10. Go through all the details of the proof of Cauchy’s theorem (Theorem 6.9 in the course notes). Justify each inequality and each equality. Summary of Week 5: • Proof of de Moivre’s theorem. • Roots of unity and other complex numbers.

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