Roots of Complex Numbers (Mα+Hs)Smart Workshop Semester 2, 2016

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((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 1 / 29 Roots of complex numbers (mα+hs)Smart Workshop Semester 2, 2016

Geoﬀ Coates

These slides describe how to ﬁnd all of the n−th roots of real and complex numbers.

Before you start, it helps to be familiar with the following topics: Representing complex numbers on the complex plane (aka the Argand plane). Working out the polar form of a complex number. de Moivre’s Theorem. π π π π Trigonometric ratios for standard ﬁrst quadrant angles ( 2 , 4 , 3 and 6 ) and using these to ﬁnd trig ratios for related angles in the other three quadrants.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 2 / 29 What can (mα+hs)Smart do for you?

Online Stuﬀ Drop-in Study Sessions presentation slides from Monday, Wednesday, Friday, workshops on many topics 10am-12pm, Ground Floor practice exercises Barry J Marshall Library, teaching weeks and study short videos breaks. and more!

Email: geoﬀ[email protected] Workshops Can’t ﬁnd what you want? See our current Got a question? Workshop Calendar for this Semester’s topics. Drop us a line!

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 3 / 29 Contents

Introduction Go

Finding roots of complex numbers Go

Properties of roots of complex numbers Go

Exercises Go Solution to Exercise 1 Go Solution to Exercise 2 Go

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 4 / 29 In the Real Number system the answer is just “2”.

Now that we have extended the number realm to include Complex Numbers, it turns out that there are two more answers to this question.

In fact, the question “what are the n−th roots of z?” will always have n answers.

Introduction

Up until now there has been only one answer to the question

“what are the cube (or third) roots of 8?”

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 5 / 29 Now that we have extended the number realm to include Complex Numbers, it turns out that there are two more answers to this question.

In fact, the question “what are the n−th roots of z?” will always have n answers.

Introduction

Up until now there has been only one answer to the question

“what are the cube (or third) roots of 8?”

In the Real Number system the answer is just “2”.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 5 / 29 In fact, the question “what are the n−th roots of z?” will always have n answers.

Introduction

Up until now there has been only one answer to the question

“what are the cube (or third) roots of 8?”

In the Real Number system the answer is just “2”.

Now that we have extended the number realm to include Complex Numbers, it turns out that there are two more answers to this question.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 5 / 29 Introduction

Up until now there has been only one answer to the question

“what are the cube (or third) roots of 8?”

In the Real Number system the answer is just “2”.

Now that we have extended the number realm to include Complex Numbers, it turns out that there are two more answers to this question.

In fact, the question “what are the n−th roots of z?” will always have n answers.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 5 / 29 Obviously, 8 is a real number but it can still be expressed as a complex number:

z = 8 + 0i

Finding roots of complex numbers

To ﬁnd the remaining two cube roots of 8 we need to convert 8 to polar form.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 6 / 29 z = 8 + 0i

Finding roots of complex numbers

To ﬁnd the remaining two cube roots of 8 we need to convert 8 to polar form.

Obviously, 8 is a real number but it can still be expressed as a complex number:

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 6 / 29 Finding roots of complex numbers

To ﬁnd the remaining two cube roots of 8 we need to convert 8 to polar form.

Obviously, 8 is a real number but it can still be expressed as a complex number:

z = 8 + 0i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 6 / 29 8.

θ is the angle the radius makes with the positive x−axis (with anticlockwise as the positive direction) so θ = 0.

imaginary (y)

−8 −6 −4 −2 2 4 6 8 real (x) −2i

−4i

−6i

−8i

Now, R is distance (or radius) of this point from (0, 0) so R =

Finding roots of complex numbers

To convert z = 8 + 0i to polar form z = R(cos θ + i sin θ), start by representing it on the complex plane:

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 7 / 29 8.

θ is the angle the radius makes with the positive x−axis (with anticlockwise as the positive direction) so θ = 0.

Now, R is distance (or radius) of this point from (0, 0) so R =

Finding roots of complex numbers

To convert z = 8 + 0i to polar form z = R(cos θ + i sin θ), start by representing it on the complex plane:

imaginary (y)

−8 −6 −4 −2 2 4 6 8 real (x) −2i

−4i

−6i

−8i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 7 / 29 θ is the angle the radius makes with the positive x−axis (with anticlockwise as the positive direction) so θ = 0.

Finding roots of complex numbers

To convert z = 8 + 0i to polar form z = R(cos θ + i sin θ), start by representing it on the complex plane:

imaginary (y)

−8 −6 −4 −2 2 4 6 8 real (x) −2i

−4i

−6i

−8i

Now, R is distance (or radius) of this point from (0, 0) so R =

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 7 / 29 8.

θ is the angle the radius makes with the positive x−axis (with anticlockwise as the positive direction) so θ = 0.

Finding roots of complex numbers

To convert z = 8 + 0i to polar form z = R(cos θ + i sin θ), start by representing it on the complex plane:

imaginary (y)

−8 −6 −4 −2 2 4 6 8 real (x) −2i

−4i

−6i

−8i

Now, R is distance (or radius) of this point from (0, 0) so R =

Finding roots of complex numbers

To convert z = 8 + 0i to polar form z = R(cos θ + i sin θ), start by representing it on the complex plane:

imaginary (y)

−8 −6 −4 −2 2 4 6 8 real (x) −2i

−4i

−6i

−8i

Now, R is distance (or radius) of this point from (0, 0) so R = 8.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 7 / 29 0.

Finding roots of complex numbers

To convert z = 8 + 0i to polar form z = R(cos θ + i sin θ), start by representing it on the complex plane:

imaginary (y)

−8 −6 −4 −2 2 4 6 8 real (x) −2i

−4i

−6i

−8i

Now, R is distance (or radius) of this point from (0, 0) so R = 8.

θ is the angle the radius makes with the positive x−axis (with anticlockwise as the positive direction) so θ =

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 7 / 29 Finding roots of complex numbers

To convert z = 8 + 0i to polar form z = R(cos θ + i sin θ), start by representing it on the complex plane:

imaginary (y)

−8 −6 −4 −2 2 4 6 8 real (x) −2i

−4i

−6i

−8i

Now, R is distance (or radius) of this point from (0, 0) so R = 8.

θ is the angle the radius makes with the positive x−axis (with anticlockwise as the positive direction) so θ = 0.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 7 / 29 w 3 = 8(cos 0 + i sin 0) in polar form 1 3 1 w 3 =[ 8(cos 0 + i sin 0)] 3 take cube root to isolate w

1 3 1 1 w = 8 cos 3 × 0 + i sin 3 × 0 by de Moivre’s Theorem = 2 (cos 0 + i sin 0) = 2(1 + i0) = 2

Great! We already knew that! So what’s the point of using polar form?

Finding roots of complex numbers

So, if w is one of the third roots of 8, we have

w 3 = 8

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 8 / 29 1 3 1 w 3 =[ 8(cos 0 + i sin 0)] 3 take cube root to isolate w

1 3 1 1 w = 8 cos 3 × 0 + i sin 3 × 0 by de Moivre’s Theorem = 2 (cos 0 + i sin 0) = 2(1 + i0) = 2

Great! We already knew that! So what’s the point of using polar form?

Finding roots of complex numbers

So, if w is one of the third roots of 8, we have

w 3 = 8 w 3 = 8(cos 0 + i sin 0) in polar form

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 8 / 29 1 3 1 1 w = 8 cos 3 × 0 + i sin 3 × 0 by de Moivre’s Theorem = 2 (cos 0 + i sin 0) = 2(1 + i0) = 2

Great! We already knew that! So what’s the point of using polar form?

Finding roots of complex numbers

So, if w is one of the third roots of 8, we have

w 3 = 8 w 3 = 8(cos 0 + i sin 0) in polar form 1 3 1 w 3 =[ 8(cos 0 + i sin 0)] 3 take cube root to isolate w

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 8 / 29 = 2 (cos 0 + i sin 0) = 2(1 + i0) = 2

Great! We already knew that! So what’s the point of using polar form?

Finding roots of complex numbers

So, if w is one of the third roots of 8, we have

w 3 = 8 w 3 = 8(cos 0 + i sin 0) in polar form 1 3 1 w 3 =[ 8(cos 0 + i sin 0)] 3 take cube root to isolate w

1 3 1 1 w = 8 cos 3 × 0 + i sin 3 × 0 by de Moivre’s Theorem

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 8 / 29 = 2(1 + i0) = 2

Great! We already knew that! So what’s the point of using polar form?

Finding roots of complex numbers

So, if w is one of the third roots of 8, we have

w 3 = 8 w 3 = 8(cos 0 + i sin 0) in polar form 1 3 1 w 3 =[ 8(cos 0 + i sin 0)] 3 take cube root to isolate w

1 3 1 1 w = 8 cos 3 × 0 + i sin 3 × 0 by de Moivre’s Theorem = 2 (cos 0 + i sin 0)

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 8 / 29 = 2

Great! We already knew that! So what’s the point of using polar form?

Finding roots of complex numbers

So, if w is one of the third roots of 8, we have

w 3 = 8 w 3 = 8(cos 0 + i sin 0) in polar form 1 3 1 w 3 =[ 8(cos 0 + i sin 0)] 3 take cube root to isolate w

1 3 1 1 w = 8 cos 3 × 0 + i sin 3 × 0 by de Moivre’s Theorem = 2 (cos 0 + i sin 0) = 2(1 + i0)

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 8 / 29 Great! We already knew that! So what’s the point of using polar form?

Finding roots of complex numbers

So, if w is one of the third roots of 8, we have

w 3 = 8 w 3 = 8(cos 0 + i sin 0) in polar form 1 3 1 w 3 =[ 8(cos 0 + i sin 0)] 3 take cube root to isolate w

1 3 1 1 w = 8 cos 3 × 0 + i sin 3 × 0 by de Moivre’s Theorem = 2 (cos 0 + i sin 0) = 2(1 + i0) = 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 8 / 29 Finding roots of complex numbers

So, if w is one of the third roots of 8, we have

w 3 = 8 w 3 = 8(cos 0 + i sin 0) in polar form 1 3 1 w 3 =[ 8(cos 0 + i sin 0)] 3 take cube root to isolate w

1 3 1 1 w = 8 cos 3 × 0 + i sin 3 × 0 by de Moivre’s Theorem = 2 (cos 0 + i sin 0) = 2(1 + i0) = 2

Great! We already knew that! So what’s the point of using polar form?

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 8 / 29 Notice that rotating the radius through a full circle (2π radians) gets us back to the same place.

In fact, rotating the radius through a full circle any number of times (k) gets us back to the same place.

Finding roots of complex numbers

Well, polar form of complex numbers includes a ﬂexibility we can use to ﬁnd other roots.

imaginary (y)

−8 −6 −4 −2 2 4 6 8 real (x) −2i

−4i

−6i

−8i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 9 / 29 In fact, rotating the radius through a full circle any number of times (k) gets us back to the same place.

Finding roots of complex numbers

Well, polar form of complex numbers includes a ﬂexibility we can use to ﬁnd other roots.

imaginary (y)

−8 −6 −4 −2 2 4 6 8 real (x) −2i

−4i

−6i

−8i

Notice that rotating the radius through a full circle (2π radians) gets us back to the same place.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 9 / 29 Finding roots of complex numbers

Well, polar form of complex numbers includes a ﬂexibility we can use to ﬁnd other roots.

imaginary (y)

−8 −6 −4 −2 2 4 6 8 real (x) −2i

−4i

−6i

−8i

Notice that rotating the radius through a full circle (2π radians) gets us back to the same place.

In fact, rotating the radius through a full circle any number of times (k) gets us back to the same place. ((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 9 / 29 Reworking the calculation of w now gives

1 w = [8(cos(0 + 2kπ) + i sin(0 + 2kπ)] 3

1 3 2kπ 2kπ = 8 cos 3 + i sin 3 by de Moivre’s Theorem

2kπ 2kπ = 2 cos 3 + i sin 3

So, diﬀerent values of k lead to diﬀerent roots.

Finding roots of complex numbers

So, technically

z = 8 = 8(cos(0 + 2kπ) + i sin(0 + 2kπ))

where k is any integer.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 10 / 29 1 3 2kπ 2kπ = 8 cos 3 + i sin 3 by de Moivre’s Theorem

2kπ 2kπ = 2 cos 3 + i sin 3

So, diﬀerent values of k lead to diﬀerent roots.

Finding roots of complex numbers

So, technically

z = 8 = 8(cos(0 + 2kπ) + i sin(0 + 2kπ))

where k is any integer.

Reworking the calculation of w now gives

1 w = [8(cos(0 + 2kπ) + i sin(0 + 2kπ)] 3

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 10 / 29 2kπ 2kπ = 2 cos 3 + i sin 3

So, diﬀerent values of k lead to diﬀerent roots.

Finding roots of complex numbers

So, technically

z = 8 = 8(cos(0 + 2kπ) + i sin(0 + 2kπ))

where k is any integer.

Reworking the calculation of w now gives

1 w = [8(cos(0 + 2kπ) + i sin(0 + 2kπ)] 3

1 3 2kπ 2kπ = 8 cos 3 + i sin 3 by de Moivre’s Theorem

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 10 / 29 So, diﬀerent values of k lead to diﬀerent roots.

Finding roots of complex numbers

So, technically

z = 8 = 8(cos(0 + 2kπ) + i sin(0 + 2kπ))

where k is any integer.

Reworking the calculation of w now gives

1 w = [8(cos(0 + 2kπ) + i sin(0 + 2kπ)] 3

1 3 2kπ 2kπ = 8 cos 3 + i sin 3 by de Moivre’s Theorem

2kπ 2kπ = 2 cos 3 + i sin 3

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 10 / 29 Finding roots of complex numbers

So, technically

z = 8 = 8(cos(0 + 2kπ) + i sin(0 + 2kπ))

where k is any integer.

Reworking the calculation of w now gives

1 w = [8(cos(0 + 2kπ) + i sin(0 + 2kπ)] 3

1 3 2kπ 2kπ = 8 cos 3 + i sin 3 by de Moivre’s Theorem

2kπ 2kπ = 2 cos 3 + i sin 3

So, diﬀerent values of k lead to diﬀerent roots.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 10 / 29 2π 2π For k = 1 we have w1 = 2 cos 3 + i sin 3 √ 1 3 = 2 − 2 + i × 2

(If you ﬁnd this step diﬃcult, see the workshop “Finding trig ratios for certain angles”.) √ = −1 + 3i √ So, −1 + 3i is also a third root of 8.

Finding roots of complex numbers

2kπ 2kπ wk = 2 cos 3 + i sin 3

We already know that k = 0 produces the root w0 = 2.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 11 / 29 √ 1 3 = 2 − 2 + i × 2

(If you ﬁnd this step diﬃcult, see the workshop “Finding trig ratios for certain angles”.) √ = −1 + 3i √ So, −1 + 3i is also a third root of 8.

Finding roots of complex numbers

2kπ 2kπ wk = 2 cos 3 + i sin 3

We already know that k = 0 produces the root w0 = 2.

2π 2π For k = 1 we have w1 = 2 cos 3 + i sin 3

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 11 / 29 (If you ﬁnd this step diﬃcult, see the workshop “Finding trig ratios for certain angles”.) √ = −1 + 3i √ So, −1 + 3i is also a third root of 8.

Finding roots of complex numbers

2kπ 2kπ wk = 2 cos 3 + i sin 3

We already know that k = 0 produces the root w0 = 2.

2π 2π For k = 1 we have w1 = 2 cos 3 + i sin 3 √ 1 3 = 2 − 2 + i × 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 11 / 29 √ = −1 + 3i √ So, −1 + 3i is also a third root of 8.

Finding roots of complex numbers

2kπ 2kπ wk = 2 cos 3 + i sin 3

We already know that k = 0 produces the root w0 = 2.

2π 2π For k = 1 we have w1 = 2 cos 3 + i sin 3 √ 1 3 = 2 − 2 + i × 2

(If you ﬁnd this step diﬃcult, see the workshop “Finding trig ratios for certain angles”.)

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 11 / 29 √ So, −1 + 3i is also a third root of 8.

Finding roots of complex numbers

2kπ 2kπ wk = 2 cos 3 + i sin 3

We already know that k = 0 produces the root w0 = 2.

2π 2π For k = 1 we have w1 = 2 cos 3 + i sin 3 √ 1 3 = 2 − 2 + i × 2

(If you ﬁnd this step diﬃcult, see the workshop “Finding trig ratios for certain angles”.) √ = −1 + 3i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 11 / 29 Finding roots of complex numbers

2kπ 2kπ wk = 2 cos 3 + i sin 3

We already know that k = 0 produces the root w0 = 2.

2π 2π For k = 1 we have w1 = 2 cos 3 + i sin 3 √ 1 3 = 2 − 2 + i × 2

(If you ﬁnd this step diﬃcult, see the workshop “Finding trig ratios for certain angles”.) √ = −1 + 3i √ So, −1 + 3i is also a third root of 8.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 11 / 29 √ 1 3 = 2 − 2 + i × − 2 √ = −1 − 3i

√ So, −1 − 3i is also a third root of 8.

Finding roots of complex numbers

4π 4π For k = 2 we have w2 = 2 cos 3 + i sin 3

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 12 / 29 √ = −1 − 3i

√ So, −1 − 3i is also a third root of 8.

Finding roots of complex numbers

4π 4π For k = 2 we have w2 = 2 cos 3 + i sin 3

√ 1 3 = 2 − 2 + i × − 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 12 / 29 √ So, −1 − 3i is also a third root of 8.

Finding roots of complex numbers

4π 4π For k = 2 we have w2 = 2 cos 3 + i sin 3

√ 1 3 = 2 − 2 + i × − 2 √ = −1 − 3i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 12 / 29 Finding roots of complex numbers

4π 4π For k = 2 we have w2 = 2 cos 3 + i sin 3

√ 1 3 = 2 − 2 + i × − 2 √ = −1 − 3i

√ So, −1 − 3i is also a third root of 8.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 12 / 29 So, the three third roots of 8 are √ √ w0 = 2, w1 = −1 + 3i and w2 = −1 − 3i

√ √ Feel free to check that (−1 + 3i)3 and (−1 − 3i)3 do indeed equal 8.

Finding roots of complex numbers

6π If you were to try k = 3 you would ﬁnd yourself working with cos and sin of 3 = 2π which gives the same root (w0 = 2) as for k = 0.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 13 / 29 √ √ Feel free to check that (−1 + 3i)3 and (−1 − 3i)3 do indeed equal 8.

Finding roots of complex numbers

6π If you were to try k = 3 you would ﬁnd yourself working with cos and sin of 3 = 2π which gives the same root (w0 = 2) as for k = 0.

So, the three third roots of 8 are √ √ w0 = 2, w1 = −1 + 3i and w2 = −1 − 3i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 13 / 29 Finding roots of complex numbers

6π If you were to try k = 3 you would ﬁnd yourself working with cos and sin of 3 = 2π which gives the same root (w0 = 2) as for k = 0.

So, the three third roots of 8 are √ √ w0 = 2, w1 = −1 + 3i and w2 = −1 − 3i

√ √ Feel free to check that (−1 + 3i)3 and (−1 − 3i)3 do indeed equal 8.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 13 / 29 Roots of real numbers occur in conjugate pairs.

√ √ In our example, both −1 + 3i and −1 − 3i are third roots of 8.

Properties of roots of complex numbers

There are some useful properties of these complex roots.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 14 / 29 √ √ In our example, both −1 + 3i and −1 − 3i are third roots of 8.

Properties of roots of complex numbers

There are some useful properties of these complex roots.

Roots of real numbers occur in conjugate pairs.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 14 / 29 Properties of roots of complex numbers

There are some useful properties of these complex roots.

Roots of real numbers occur in conjugate pairs.

√ √ In our example, both −1 + 3i and −1 − 3i are third roots of 8.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 14 / 29 Notice that the radius of all three roots is 2. This means that they all sit on a circle of radius 2 centred at the origin.

Also, the roots are evenly spaced around the circumference. The angle between each 2π radius is 3 .

Properties of roots of complex numbers

Plotting these three roots on the complex plane reveals a very useful pattern:

imaginary (y)

2i w1 √ 3i

w0 −2 −1 1 2 real (x)

−i

√ − 3i w2 −2i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 15 / 29 This means that they all sit on a circle of radius 2 centred at the origin.

Also, the roots are evenly spaced around the circumference. The angle between each 2π radius is 3 .

Properties of roots of complex numbers

Plotting these three roots on the complex plane reveals a very useful pattern:

imaginary (y)

2i w1 √ 3i

w0 −2 −1 1 2 real (x)

−i

√ − 3i w2 −2i

Notice that the radius of all three roots is 2.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 15 / 29 Also, the roots are evenly spaced around the circumference. The angle between each 2π radius is 3 .

Properties of roots of complex numbers

Plotting these three roots on the complex plane reveals a very useful pattern:

imaginary (y)

2i w1 √ 3i

w0 −2 −1 1 2 real (x)

−i

√ − 3i w2 −2i

Notice that the radius of all three roots is 2. This means that they all sit on a circle of radius 2 centred at the origin.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 15 / 29 The angle between each 2π radius is 3 .

Properties of roots of complex numbers

Plotting these three roots on the complex plane reveals a very useful pattern:

imaginary (y)

2i w1 √ 3i

w0 −2 −1 1 2 real (x)

−i

√ − 3i w2 −2i

Notice that the radius of all three roots is 2. This means that they all sit on a circle of radius 2 centred at the origin.

Also, the roots are evenly spaced around the circumference.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 15 / 29 Properties of roots of complex numbers

Plotting these three roots on the complex plane reveals a very useful pattern:

imaginary (y)

2i w1 √ 3i

w0 −2 −1 1 2 real (x)

−i

√ − 3i w2 −2i

Notice that the radius of all three roots is 2. This means that they all sit on a circle of radius 2 centred at the origin.

Also, the roots are evenly spaced around the circumference. The angle between each 2π radius is 3 . ((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 15 / 29 For example, you can now see why complex roots of real numbers occur in conjugate pairs.

Properties of roots of complex numbers

Plotting these three roots on the complex plane reveals a very useful pattern:

imaginary (y)

2i w1 √ 3i

w0 −2 −1 1 2 real (x)

−i

√ − 3i w2 −2i

These are general properties of the roots of complex numbers.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 16 / 29 Properties of roots of complex numbers

Plotting these three roots on the complex plane reveals a very useful pattern:

imaginary (y)

2i w1 √ 3i

w0 −2 −1 1 2 real (x)

−i

√ − 3i w2 −2i

These are general properties of the roots of complex numbers.

For example, you can now see why complex roots of real numbers occur in conjugate pairs.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 16 / 29 Roots of complex numbers: Exercises

Have a go at the following practice exercises and then check the worked answers on the following slides.

√ Exercise 1: Find the square roots of 2 + 2 3i.

Exercise 2: Find the 4th roots of −16.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 17 / 29 θ

q √ √ R = 22 + (2 3)2 = 16 = 4.

2 1 π From the triangle above, cos(θ) = = (or you can use sine or tan): θ = + 2kπ. 4 2 3

Solution to Exercise 1 √ Start by converting z = 2 + 2 3i to general polar form.

imaginary (y)

4i √ 2 3i−→ 3i

−4 −3 −2 −1 1 2 3 4 real (x) −i

−2i

−3i

−4i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 18 / 29 θ

2 1 π From the triangle above, cos(θ) = = (or you can use sine or tan): θ = + 2kπ. 4 2 3

Solution to Exercise 1 √ Start by converting z = 2 + 2 3i to general polar form.

imaginary (y)

4i √ 2 3i−→ 3i

2i R

−4 −3 −2 −1 1 2 3 4 real (x) −i

−2i

−3i

−4i

q √ √ R = 22 + (2 3)2 = 16 = 4.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 18 / 29 θ

2 1 π cos(θ) = = (or you can use sine or tan): θ = + 2kπ. 4 2 3

Solution to Exercise 1 √ Start by converting z = 2 + 2 3i to general polar form.

imaginary (y)

4i √ 2 3i−→ 3i

2i R

−4 −3 −2 −1 1 2 3 4 real (x) −i

−2i

−3i

−4i

q √ √ R = 22 + (2 3)2 = 16 = 4.

From the triangle above,

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 18 / 29 π (or you can use sine or tan): θ = + 2kπ. 3

Solution to Exercise 1 √ Start by converting z = 2 + 2 3i to general polar form.

imaginary (y)

4i √ 2 3i−→ 3i

2i 4

i θ −4 −3 −2 −1 1 2 3 4 real (x) −i

−2i

−3i

−4i

q √ √ R = 22 + (2 3)2 = 16 = 4.

2 1 From the triangle above, cos(θ) = = 4 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 18 / 29 π θ = + 2kπ. 3

Solution to Exercise 1 √ Start by converting z = 2 + 2 3i to general polar form.

imaginary (y)

4i √ 2 3i−→ 3i

2i 4

i θ −4 −3 −2 −1 1 2 3 4 real (x) −i

−2i

−3i

−4i

q √ √ R = 22 + (2 3)2 = 16 = 4.

2 1 From the triangle above, cos(θ) = = (or you can use sine or tan): 4 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 18 / 29 Solution to Exercise 1 √ Start by converting z = 2 + 2 3i to general polar form.

imaginary (y)

4i √ 2 3i−→ 3i

2i 4

i θ −4 −3 −2 −1 1 2 3 4 real (x) −i

−2i

−3i

−4i

q √ √ R = 22 + (2 3)2 = 16 = 4.

2 1 π From the triangle above, cos(θ) = = (or you can use sine or tan): θ = + 2kπ. 4 2 3

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 18 / 29 1 π π 2 wk = 4 cos 3 + 2kπ + i sin 3 + 2kπ

1 2 π π = 4 cos 6 + kπ + i sin 6 + kπ by de Moivre’s Theorem

π π = 2 cos 6 + kπ + i sin 6 + kπ

Solution to Exercise 1

√ π π z = 2 + 2 3i = 4 cos 3 + 2kπ + i sin 3 + 2kπ where k is any integer.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 19 / 29 1 2 π π = 4 cos 6 + kπ + i sin 6 + kπ by de Moivre’s Theorem

π π = 2 cos 6 + kπ + i sin 6 + kπ

Solution to Exercise 1

√ π π z = 2 + 2 3i = 4 cos 3 + 2kπ + i sin 3 + 2kπ where k is any integer.

1 π π 2 wk = 4 cos 3 + 2kπ + i sin 3 + 2kπ

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 19 / 29 π π = 2 cos 6 + kπ + i sin 6 + kπ

Solution to Exercise 1

√ π π z = 2 + 2 3i = 4 cos 3 + 2kπ + i sin 3 + 2kπ where k is any integer.

1 π π 2 wk = 4 cos 3 + 2kπ + i sin 3 + 2kπ

1 2 π π = 4 cos 6 + kπ + i sin 6 + kπ by de Moivre’s Theorem

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 19 / 29 Solution to Exercise 1

√ π π z = 2 + 2 3i = 4 cos 3 + 2kπ + i sin 3 + 2kπ where k is any integer.

1 π π 2 wk = 4 cos 3 + 2kπ + i sin 3 + 2kπ

1 2 π π = 4 cos 6 + kπ + i sin 6 + kπ by de Moivre’s Theorem

π π = 2 cos 6 + kπ + i sin 6 + kπ

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 19 / 29 √ 3 1 = 2 2 + i × 2 √ = 3 + i

7π 7π For k = 1 we have w1 = 2 cos 6 + i sin 6

√ 3 1 = 2 − 2 − i × 2 √ = − 3 − i

π Note: Making k = 2 produces the angle 6 + 2π which gives the same root as for k = 0.

Solution to Exercise 1

π π wk = 2 cos 6 + kπ + i sin 6 + kπ

π π For k = 0 we have w0 = 2 cos 6 + i sin 6

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 20 / 29 √ = 3 + i

7π 7π For k = 1 we have w1 = 2 cos 6 + i sin 6

√ 3 1 = 2 − 2 − i × 2 √ = − 3 − i

π Note: Making k = 2 produces the angle 6 + 2π which gives the same root as for k = 0.

Solution to Exercise 1

π π wk = 2 cos 6 + kπ + i sin 6 + kπ

π π For k = 0 we have w0 = 2 cos 6 + i sin 6

√ 3 1 = 2 2 + i × 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 20 / 29 7π 7π For k = 1 we have w1 = 2 cos 6 + i sin 6

√ 3 1 = 2 − 2 − i × 2 √ = − 3 − i

π Note: Making k = 2 produces the angle 6 + 2π which gives the same root as for k = 0.

Solution to Exercise 1

π π wk = 2 cos 6 + kπ + i sin 6 + kπ

π π For k = 0 we have w0 = 2 cos 6 + i sin 6

√ 3 1 = 2 2 + i × 2 √ = 3 + i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 20 / 29 √ 3 1 = 2 − 2 − i × 2 √ = − 3 − i

π Note: Making k = 2 produces the angle 6 + 2π which gives the same root as for k = 0.

Solution to Exercise 1

π π wk = 2 cos 6 + kπ + i sin 6 + kπ

π π For k = 0 we have w0 = 2 cos 6 + i sin 6

√ 3 1 = 2 2 + i × 2 √ = 3 + i

7π 7π For k = 1 we have w1 = 2 cos 6 + i sin 6

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 20 / 29 √ = − 3 − i

π Note: Making k = 2 produces the angle 6 + 2π which gives the same root as for k = 0.

Solution to Exercise 1

π π wk = 2 cos 6 + kπ + i sin 6 + kπ

π π For k = 0 we have w0 = 2 cos 6 + i sin 6

√ 3 1 = 2 2 + i × 2 √ = 3 + i

7π 7π For k = 1 we have w1 = 2 cos 6 + i sin 6

√ 3 1 = 2 − 2 − i × 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 20 / 29 π Note: Making k = 2 produces the angle 6 + 2π which gives the same root as for k = 0.

Solution to Exercise 1

π π wk = 2 cos 6 + kπ + i sin 6 + kπ

π π For k = 0 we have w0 = 2 cos 6 + i sin 6

√ 3 1 = 2 2 + i × 2 √ = 3 + i

7π 7π For k = 1 we have w1 = 2 cos 6 + i sin 6

√ 3 1 = 2 − 2 − i × 2 √ = − 3 − i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 20 / 29 Solution to Exercise 1

π π wk = 2 cos 6 + kπ + i sin 6 + kπ

π π For k = 0 we have w0 = 2 cos 6 + i sin 6

√ 3 1 = 2 2 + i × 2 √ = 3 + i

7π 7π For k = 1 we have w1 = 2 cos 6 + i sin 6

√ 3 1 = 2 − 2 − i × 2 √ = − 3 − i

π Note: Making k = 2 produces the angle 6 + 2π which gives the same root as for k = 0.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 20 / 29 √ √ √ Feel free to check that ( 3 + i)2 and (− 3 − i)2 both equal 2 + 2 3i.

Note: Since these are roots of a complex number they do not occur in a conjugate pair.

Solution to Exercise 1

√ So, the two second (ie. square) roots of z = 2 + 2 3i are √ √ w0 = 3 + i and w1 = − 3 − i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 21 / 29 Note: Since these are roots of a complex number they do not occur in a conjugate pair.

Solution to Exercise 1

√ So, the two second (ie. square) roots of z = 2 + 2 3i are √ √ w0 = 3 + i and w1 = − 3 − i

√ √ √ Feel free to check that ( 3 + i)2 and (− 3 − i)2 both equal 2 + 2 3i.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 21 / 29 Solution to Exercise 1

√ So, the two second (ie. square) roots of z = 2 + 2 3i are √ √ w0 = 3 + i and w1 = − 3 − i

√ √ √ Feel free to check that ( 3 + i)2 and (− 3 − i)2 both equal 2 + 2 3i.

Note: Since these are roots of a complex number they do not occur in a conjugate pair.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 21 / 29 The radius of both roots is 2.

The roots are evenly spaced around the circle, separated by an angle of π.

Solution to Exercise 1

Plot these two roots on the complex plane to check the “wagon wheel” pattern:

imaginary (y) 4i z 3i

2i w0 i

−4 −3 −2 −1 1 2 3 4 real (x) −i w1 −2i

−3i

−4i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 22 / 29 The roots are evenly spaced around the circle, separated by an angle of π.

Solution to Exercise 1

Plot these two roots on the complex plane to check the “wagon wheel” pattern:

imaginary (y) 4i z 3i

2i w0 i

−4 −3 −2 −1 1 2 3 4 real (x) −i w1 −2i

−3i

−4i

The radius of both roots is 2.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 22 / 29 Solution to Exercise 1

Plot these two roots on the complex plane to check the “wagon wheel” pattern:

imaginary (y) 4i z 3i

2i w0 i

−4 −3 −2 −1 1 2 3 4 real (x) −i w1 −2i

−3i

−4i

The radius of both roots is 2.

The roots are evenly spaced around the circle, separated by an angle of π.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 22 / 29 R = p(−16)2 + 02 = 16.

θ = π + 2kπ.

Solution to Exercise 2

Start by converting z = −16 = −16 + 0i to general polar form.

imaginary (y)

16i

12i

−16 −12 −8 −4 4 8 12 16 real (x) −4i

−8i

−12i

−16i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 23 / 29 θ = π + 2kπ.

Solution to Exercise 2

Start by converting z = −16 = −16 + 0i to general polar form.

imaginary (y)

16i

12i

−16 −12 −8 −4 4 8 12 16 real (x) −4i

−8i

−12i

−16i

R = p(−16)2 + 02 = 16.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 23 / 29 Solution to Exercise 2

Start by converting z = −16 = −16 + 0i to general polar form.

imaginary (y)

16i

12i

−16 −12 −8 −4 4 8 12 16 real (x) −4i

−8i

−12i

−16i

R = p(−16)2 + 02 = 16.

θ = π + 2kπ.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 23 / 29 1 wk = [16 (cos (π + 2kπ) + i sin (π + 2kπ))] 4

1 4 π kπ π kπ = 16 cos 4 + 2 + i sin 4 + 2 by de Moivre’s Theorem

π kπ π kπ = 2 cos 4 + 2 + i sin 4 + 2

Solution to Exercise 2

z = −16 = 2 (cos (π + 2kπ) + i sin (π + 2kπ))

where k is any integer.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 24 / 29 1 4 π kπ π kπ = 16 cos 4 + 2 + i sin 4 + 2 by de Moivre’s Theorem

π kπ π kπ = 2 cos 4 + 2 + i sin 4 + 2

Solution to Exercise 2

z = −16 = 2 (cos (π + 2kπ) + i sin (π + 2kπ))

where k is any integer.

1 wk = [16 (cos (π + 2kπ) + i sin (π + 2kπ))] 4

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 24 / 29 π kπ π kπ = 2 cos 4 + 2 + i sin 4 + 2

Solution to Exercise 2

z = −16 = 2 (cos (π + 2kπ) + i sin (π + 2kπ))

where k is any integer.

1 wk = [16 (cos (π + 2kπ) + i sin (π + 2kπ))] 4

1 4 π kπ π kπ = 16 cos 4 + 2 + i sin 4 + 2 by de Moivre’s Theorem

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 24 / 29 Solution to Exercise 2

z = −16 = 2 (cos (π + 2kπ) + i sin (π + 2kπ))

where k is any integer.

1 wk = [16 (cos (π + 2kπ) + i sin (π + 2kπ))] 4

1 4 π kπ π kπ = 16 cos 4 + 2 + i sin 4 + 2 by de Moivre’s Theorem

π kπ π kπ = 2 cos 4 + 2 + i sin 4 + 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 24 / 29 √ √ 2 2 = 2 2 + i × 2 √ √ = 2 + 2i

3π 3π For k = 1 we have w1 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 − 2 + i × 2 √ √ = − 2 + 2i

Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

π π For k = 0 we have w0 = 2 cos 4 + i sin 4

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 25 / 29 √ √ = 2 + 2i

3π 3π For k = 1 we have w1 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 − 2 + i × 2 √ √ = − 2 + 2i

Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

π π For k = 0 we have w0 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 2 + i × 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 25 / 29 3π 3π For k = 1 we have w1 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 − 2 + i × 2 √ √ = − 2 + 2i

Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

π π For k = 0 we have w0 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 2 + i × 2 √ √ = 2 + 2i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 25 / 29 √ √ 2 2 = 2 − 2 + i × 2 √ √ = − 2 + 2i

Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

π π For k = 0 we have w0 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 2 + i × 2 √ √ = 2 + 2i

3π 3π For k = 1 we have w1 = 2 cos 4 + i sin 4

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 25 / 29 √ √ = − 2 + 2i

Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

π π For k = 0 we have w0 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 2 + i × 2 √ √ = 2 + 2i

3π 3π For k = 1 we have w1 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 − 2 + i × 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 25 / 29 Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

π π For k = 0 we have w0 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 2 + i × 2 √ √ = 2 + 2i

3π 3π For k = 1 we have w1 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 − 2 + i × 2 √ √ = − 2 + 2i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 25 / 29 √ √ 2 2 = 2 − 2 − i × 2 √ √ = − 2 − 2i

7π 7π For k = 3 we have w3 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 2 − i × 2 √ √ = 2 − 2i

Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

5π 5π For k = 2 we have w2 = 2 cos 4 + i sin 4

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 26 / 29 √ √ = − 2 − 2i

7π 7π For k = 3 we have w3 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 2 − i × 2 √ √ = 2 − 2i

Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

5π 5π For k = 2 we have w2 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 − 2 − i × 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 26 / 29 7π 7π For k = 3 we have w3 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 2 − i × 2 √ √ = 2 − 2i

Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

5π 5π For k = 2 we have w2 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 − 2 − i × 2 √ √ = − 2 − 2i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 26 / 29 √ √ 2 2 = 2 2 − i × 2 √ √ = 2 − 2i

Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

5π 5π For k = 2 we have w2 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 − 2 − i × 2 √ √ = − 2 − 2i

7π 7π For k = 3 we have w3 = 2 cos 4 + i sin 4

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 26 / 29 √ √ = 2 − 2i

Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

5π 5π For k = 2 we have w2 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 − 2 − i × 2 √ √ = − 2 − 2i

7π 7π For k = 3 we have w3 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 2 − i × 2

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 26 / 29 Solution to Exercise 2

π kπ π kπ wk = 2 cos 4 + 2 + i sin 4 + 2

5π 5π For k = 2 we have w2 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 − 2 − i × 2 √ √ = − 2 − 2i

7π 7π For k = 3 we have w3 = 2 cos 4 + i sin 4

√ √ 2 2 = 2 2 − i × 2 √ √ = 2 − 2i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 26 / 29 Note: w0 and w3 are a conjugate pair. So are w1 and w2.

Solution to Exercise 2

So, the four fourth roots of z = −16 are √ √ √ √ √ √ √ √ w0 = 2 + 2i, w1 = − 2 + 2i, w2 = − 2 − 2i, and w3 = 2 − 2i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 27 / 29 So are w1 and w2.

Solution to Exercise 2

So, the four fourth roots of z = −16 are √ √ √ √ √ √ √ √ w0 = 2 + 2i, w1 = − 2 + 2i, w2 = − 2 − 2i, and w3 = 2 − 2i

Note: w0 and w3 are a conjugate pair.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 27 / 29 Solution to Exercise 2

So, the four fourth roots of z = −16 are √ √ √ √ √ √ √ √ w0 = 2 + 2i, w1 = − 2 + 2i, w2 = − 2 − 2i, and w3 = 2 − 2i

Note: w0 and w3 are a conjugate pair. So are w1 and w2.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 27 / 29 The radius of all four roots is 2.

π The roots are evenly spaced around the circle, separated by an angle of 2 .

Solution to Exercise 2

Plot the roots on the complex plane to check the “wagon wheel” pattern:

imaginary (y)

2i w1 w0 ←− z is way down here i ←− z is way down here

−4 −3 −2 −1 1 2 3 4 real (x) −i

w2 −2i w3

−3i

−4i

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 28 / 29 π The roots are evenly spaced around the circle, separated by an angle of 2 .

Solution to Exercise 2

Plot the roots on the complex plane to check the “wagon wheel” pattern:

imaginary (y)

2i w1 w0 ←− z is way down here i ←− z is way down here

−4 −3 −2 −1 1 2 3 4 real (x) −i

w2 −2i w3

−3i

−4i

The radius of all four roots is 2.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 28 / 29 Solution to Exercise 2

Plot the roots on the complex plane to check the “wagon wheel” pattern:

imaginary (y)

2i w1 w0 ←− z is way down here i ←− z is way down here

−4 −3 −2 −1 1 2 3 4 real (x) −i

w2 −2i w3

−3i

−4i

The radius of all four roots is 2.

π The roots are evenly spaced around the circle, separated by an angle of 2 .

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 28 / 29 Using STUDYSmarter Resources

This resource was developed for UWA students by the STUDYSmarter team for the numeracy program. When using our resources, please retain them in their original form with both the STUDYSmarter heading and the UWA crest.

((mα+hs)SmartRootsWorkshop of complexSemester numbers 2, 2016) Contents Prev Next 29 / 29