De Moivre's Theorem — Mathslive.ie

§1

Where De Moivre's comes from

We already know how to multiply two complex numbers in polar form.

$z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$

$z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$

$z_1 z_2 = r_1 r_2\bigl(\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)\bigr)$

Moduli multiply. Arguments add.

Now watch what happens when we square a complex number — just use the rule with $z_1 = z_2 = z$.

$z^2 = r \cdot r \bigl(\cos(\theta + \theta) + i\sin(\theta + \theta)\bigr)$

$z^2 = r^2(\cos 2\theta + i\sin 2\theta)$

And cubing — that's $z^2 \cdot z$.

$z^3 = r^2 \cdot r\bigl(\cos(2\theta + \theta) + i\sin(2\theta + \theta)\bigr)$

$z^3 = r^3(\cos 3\theta + i\sin 3\theta)$

See the pattern? The modulus picks up the power. The angle gets multiplied by the power.

De Moivre's Theorem — must learn

If $z = r(\cos\theta + i\sin\theta)$, then for any integer $n$:

$z^n = r^n(\cos n\theta + i\sin n\theta)$

In words: raise the modulus to the power, multiply the angle by the power. That's it. The whole theorem.

§2

Whole powers — Application 1

The hard way: multiply $(1 + \sqrt{3}\,i)$ by itself ten times. The smart way: polar form + De Moivre's.

(i) If $z = 1 + \sqrt{3}\,i$, find $z^{10}$.

Step 1 — find $r$ and $\theta$.

$r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$

$\tan\theta = \dfrac{\sqrt{3}}{1} \;\Rightarrow\; \theta = 60^\circ$

$z = 2(\cos 60^\circ + i\sin 60^\circ)$

Step 2 — apply De Moivre's with $n = 10$.

$z^{10} = 2^{10}\bigl(\cos(10 \cdot 60^\circ) + i\sin(10 \cdot 60^\circ)\bigr)$

$z^{10} = 2^{10}(\cos 600^\circ + i\sin 600^\circ)$

$600^\circ$ is more than $360^\circ$ — knock off a full revolution: $600 - 360 = 240^\circ$.

$\cos 240^\circ = -\dfrac{1}{2}, \quad \sin 240^\circ = -\dfrac{\sqrt{3}}{2}$

$z^{10} = 2^{10}\left(-\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\,i\right)$

$z^{10} = \dfrac{2^{10}}{2}\bigl(-1 - \sqrt{3}\,i\bigr)$

$z^{10} = -2^{9}(1 + \sqrt{3}\,i)$

Leave it in factored form. $2^9 = 512$ if you want a number, but $-2^9(1 + \sqrt{3}\,i)$ is cleaner.

Try-me 1

If $z = 1 + i$, find $z^{8}$.

Find $r$ and $\theta$ first. Then De Moivre's with $n = 8$.

$r = \sqrt{1 + 1} = \sqrt{2}, \quad \theta = 45^\circ$

$z = \sqrt{2}(\cos 45^\circ + i\sin 45^\circ)$

$z^{8} = (\sqrt{2})^{8}(\cos 360^\circ + i\sin 360^\circ)$

$= 16(1 + 0i)$

$z^{8} = 16$

Try-me 2

If $z = -1 + i$, find $z^{6}$.

$z$ is in the second quadrant — angle is $135^\circ$, not $45^\circ$.

$r = \sqrt{1 + 1} = \sqrt{2}, \quad \theta = 135^\circ$

$z^{6} = (\sqrt{2})^{6}(\cos 810^\circ + i\sin 810^\circ)$

$810^\circ - 720^\circ = 90^\circ$

$= 8(\cos 90^\circ + i\sin 90^\circ)$

$= 8(0 + i)$

$z^{6} = 8i$

Try-me 3

If $z = \sqrt{3} - i$, find $z^{5}$ in the form $a + bi$.

$z$ is in the fourth quadrant — angle is $-30^\circ$ (or $330^\circ$).

$r = \sqrt{3 + 1} = 2, \quad \theta = -30^\circ$

$z^{5} = 2^{5}\bigl(\cos(-150^\circ) + i\sin(-150^\circ)\bigr)$

$= 32\left(-\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i\right)$

$z^{5} = -16\sqrt{3} - 16i$

§3

The Argand picture

Each time you raise $z$ to a power, two things happen on the Argand diagram:

• The point rotates by $\theta$ (the angle gets added each time).

• The distance from the origin becomes $r^n$.

(i) If $z = \sqrt{3} + i$, plot $z$, $z^{2}$ and $z^{3}$.

$r = \sqrt{3 + 1} = 2, \quad \theta = 30^\circ$

$z = 2(\cos 30^\circ + i\sin 30^\circ) = \sqrt{3} + i$

$z^{2} = 4(\cos 60^\circ + i\sin 60^\circ) = 2 + 2\sqrt{3}\,i$

$z^{3} = 8(\cos 90^\circ + i\sin 90^\circ) = 0 + 8i$

On the Argand diagram, $z$, $z^{2}$, $z^{3}$ spiral outwards — each one further from the origin and rotated $30^\circ$ further round.

Three cases — depending on whether the modulus is bigger than $1$, smaller than $1$, or exactly $1$:

What $r^{n}$ does

(i)$r > 1$ ⇒ further from origin each power. $r = 2$ gives $r^2 = 4,\ r^3 = 8,\ \ldots$ — spirals out.

(ii)$r < 1$ ⇒ closer to origin each power. $r = \tfrac{1}{2}$ gives $r^2 = \tfrac{1}{4},\ r^3 = \tfrac{1}{8},\ \ldots$ — spirals in.

(iii)$r = 1$ ⇒ stays on the unit circle. $r^n = 1$ for every $n$. Pure rotation.

All three cases at a glance — same starting angle, three different moduli:

(ii) If $z = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\,i$, find $z^{3}$.

First, $r$ and $\theta$.

$r = \sqrt{\tfrac{1}{4} + \tfrac{3}{4}} = 1$

This is on the unit circle — we expect $z^3$ to stay on the unit circle too.

$z$ is in the third quadrant. $\theta = 240^\circ$ (or equivalently $-120^\circ$).

$z = \cos 240^\circ + i\sin 240^\circ$

$z^{3} = (\cos 240^\circ + i\sin 240^\circ)^{3}$

$z^{3} = \cos 720^\circ + i\sin 720^\circ$

$720^\circ$ is two full revolutions — we're back at $0^\circ$.

$z^{3} = 1 + 0i = 1$

Picture it on the unit circle — each power rotates by $240^\circ$. After three rotations we've gone $720^\circ$, exactly two full revolutions, and we're back at $1$.

Try-me 4

If $z = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\,i$, find $z^{6}$.

$r = 1$ — should land somewhere on the unit circle.

$r = \sqrt{\tfrac{1}{4} + \tfrac{3}{4}} = 1, \quad \theta = 60^\circ$

$z^{6} = 1^{6}(\cos 360^\circ + i\sin 360^\circ)$

$= 1 + 0i$

$z^{6} = 1$

Try-me 5

If $z = \dfrac{1}{2}(\sqrt{3} + i)$, find $z^{4}$ in the form $a + bi$.

Multiply out first: $z = \dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i$. Then $r$ and $\theta$.

$z = \dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i$

$r = \sqrt{\tfrac{3}{4} + \tfrac{1}{4}} = 1, \quad \theta = 30^\circ$

$z^{4} = 1^{4}(\cos 120^\circ + i\sin 120^\circ)$

$= -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$

$z^{4} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\,i$

§4

Fractional powers — Application 2

For whole powers, the polar form $r(\cos\theta + i\sin\theta)$ is fine — every power gives one answer.

But the equation $z^2 = 2 + 2\sqrt{3}\,i$ should have two answers, and $z^4 = \ldots$ should have four. To get them all we need the general polar form — adding multiples of $360^\circ$ to the angle, because cos and sin repeat every $360^\circ$.

Critical rule

•Whole power ⇒ use polar form $r(\cos\theta + i\sin\theta)$.

•Fractional power ⇒ use general polar form $r\bigl(\cos(\theta + 360n) + i\sin(\theta + 360n)\bigr)$, with $n = 0, 1, 2, \ldots$

The number of roots = the denominator of the fractional power. Square root: 2 values. Cube root: 3 values. Fourth root: 4 values.

(i) If $z^{2} = 2 + 2\sqrt{3}\,i$, find the 2 values of $z$.

$z = (2 + 2\sqrt{3}\,i)^{1/2}$

Put the right-hand side into polar form.

$r = \sqrt{4 + 12} = 4, \quad \theta = 60^\circ$

$2 + 2\sqrt{3}\,i = 4(\cos 60^\circ + i\sin 60^\circ)$

Fractional power — switch to general polar form.

$2 + 2\sqrt{3}\,i = 4\bigl(\cos(60 + 360n) + i\sin(60 + 360n)\bigr)$

$z = (2 + 2\sqrt{3}\,i)^{1/2} = 4^{1/2}\left[\cos\tfrac{1}{2}(60 + 360n) + i\sin\tfrac{1}{2}(60 + 360n)\right]$

$z = 2\bigl[\cos(30 + 180n) + i\sin(30 + 180n)\bigr]$

Now sub in $n = 0, 1$ — that's two values, which is what we need.

$n = 0:\ \ z = 2(\cos 30^\circ + i\sin 30^\circ) = 2\left(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i\right) = \sqrt{3} + i$

$n = 1:\ \ z = 2(\cos 210^\circ + i\sin 210^\circ) = 2\left(-\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i\right) = -\sqrt{3} - i$

$z = \sqrt{3} + i \quad \text{or} \quad z = -\sqrt{3} - i$

Notice: the two values differ by a sign. That's because going round $180^\circ$ flips the point through the origin.

Why not $n = 2$? That would give angle $390^\circ = 30^\circ$ — same as $n = 0$. After $n = 0, 1$ the values just repeat.

(ii) Find the four complex numbers $z$ such that $z^{4} = -8 - 8\sqrt{3}\,i$.

$z = (-8 - 8\sqrt{3}\,i)^{1/4}$

Polar form of the right-hand side.

$r = \sqrt{64 + 192} = \sqrt{256} = 16$

$-8 - 8\sqrt{3}\,i$ is in the third quadrant. $\theta = 240^\circ$.

$-8 - 8\sqrt{3}\,i = 16(\cos 240^\circ + i\sin 240^\circ)$

Fractional power — general polar form.

$-8 - 8\sqrt{3}\,i = 16\bigl(\cos(240 + 360n) + i\sin(240 + 360n)\bigr)$

$z = 16^{1/4}\left[\cos\tfrac{1}{4}(240 + 360n) + i\sin\tfrac{1}{4}(240 + 360n)\right]$

$z = 2\bigl[\cos(60 + 90n) + i\sin(60 + 90n)\bigr]$

Now sub $n = 0, 1, 2, 3$.

$n = 0:\ \ z = 2(\cos 60^\circ + i\sin 60^\circ) = 2\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right) = 1 + \sqrt{3}\,i$

$n = 1:\ \ z = 2(\cos 150^\circ + i\sin 150^\circ) = 2\left(-\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i\right) = -\sqrt{3} + i$

$n = 2:\ \ z = 2(\cos 240^\circ + i\sin 240^\circ) = 2\left(-\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i\right) = -1 - \sqrt{3}\,i$

$n = 3:\ \ z = 2(\cos 330^\circ + i\sin 330^\circ) = 2\left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i\right) = \sqrt{3} - i$

$z \in \{\, 1 + \sqrt{3}\,i,\ \ -\sqrt{3} + i,\ \ -1 - \sqrt{3}\,i,\ \ \sqrt{3} - i \,\}$

$n = 4$ gives angle $420^\circ = 60^\circ$ — back to $n = 0$. Always stop when you have the right number of roots.

On the Argand diagram, all four roots sit on a circle of radius $2$, equally spaced — each one $90^\circ$ apart from the next.

Try-me 6

Find the two values of $z$ such that $z^{2} = 2i$.

$2i$ has $r = 2$, $\theta = 90^\circ$. Use general polar form and try $n = 0, 1$.

$2i = 2(\cos 90^\circ + i\sin 90^\circ) = 2(\cos(90 + 360n) + i\sin(90 + 360n))$

$z = (2i)^{1/2} = 2^{1/2}\bigl[\cos(45 + 180n) + i\sin(45 + 180n)\bigr]$

$n = 0:\ \ z = \sqrt{2}\left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i\right) = 1 + i$

$n = 1:\ \ z = \sqrt{2}(\cos 225^\circ + i\sin 225^\circ) = -1 - i$

$z = 1 + i$ or $z = -1 - i$

Try-me 7

Find the three values of $z$ such that $z^{3} = 8i$, in the form $a + bi$.

$8i = 8(\cos 90^\circ + i\sin 90^\circ)$. Cube root ⇒ three values. Try $n = 0, 1, 2$.

$8i = 8(\cos(90 + 360n) + i\sin(90 + 360n))$

$z = 8^{1/3}\bigl[\cos(30 + 120n) + i\sin(30 + 120n)\bigr]$

$z = 2\bigl[\cos(30 + 120n) + i\sin(30 + 120n)\bigr]$

$n = 0:\ \ 2(\cos 30^\circ + i\sin 30^\circ) = \sqrt{3} + i$

$n = 1:\ \ 2(\cos 150^\circ + i\sin 150^\circ) = -\sqrt{3} + i$

$n = 2:\ \ 2(\cos 270^\circ + i\sin 270^\circ) = -2i$

$z = \sqrt{3} + i,\ \ -\sqrt{3} + i,\ \ -2i$

§5

Cube roots of unity — $\omega$

A special, very famous example: solve $z^{3} = 1$.

(i) Solve $z^{3} = 1$ for the three values of $z$.

$z = 1^{1/3}$

Write $1$ in general polar form. $r = 1$, $\theta = 0^\circ$.

$1 = \cos(0 + 360n) + i\sin(0 + 360n)$

$1 = \cos 360n + i\sin 360n$

$z = 1^{1/3} = \cos 120n + i\sin 120n$

$n = 0:\ \ z = \cos 0^\circ + i\sin 0^\circ = 1$

$n = 1:\ \ z = \cos 120^\circ + i\sin 120^\circ = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\,i$

$n = 2:\ \ z = \cos 240^\circ + i\sin 240^\circ = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\,i$

The two non-real ones get the special name $\omega$ (omega) and $\omega^{2}$.

The cube roots of unity

•$1, \ \omega, \ \omega^{2}$ where $\omega = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\,i$ and $\omega^{2} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\,i$

•$\omega^{3} = 1$ (because $\omega$ is a cube root of $1$)

Three points equally spaced $120^\circ$ apart on the unit circle. That symmetry leads to two essential properties.

(ii) Prove that $1 + \omega + \omega^{2} = 0$.

$1 + \omega + \omega^{2} = 1 + \left(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\,i\right) + \left(-\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\,i\right)$

$= 1 - \dfrac{1}{2} - \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\,i - \dfrac{\sqrt{3}}{2}\,i$

$= 1 - 1 + 0 = 0$ ✓

The imaginary parts cancel because $\omega^{2}$ is the conjugate of $\omega$. The real parts give $1 - \tfrac{1}{2} - \tfrac{1}{2} = 0$.

(iii) Verify $\omega^{2}$ by direct squaring.

We got $\omega^{2} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\,i$ from the polar-form route (sub $n = 2$). It must also come out by squaring $\omega$ in rectangular form. Let's check.

$\omega^{2} = \left(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\,i\right)^{2}$

Using $(a + b)^{2} = a^{2} + 2ab + b^{2}$ with $a = -\tfrac{1}{2}$ and $b = \tfrac{\sqrt{3}}{2}\,i$.

$= \left(-\dfrac{1}{2}\right)^{2} + 2\left(-\dfrac{1}{2}\right)\left(\dfrac{\sqrt{3}}{2}\,i\right) + \left(\dfrac{\sqrt{3}}{2}\,i\right)^{2}$

$= \dfrac{1}{4} - \dfrac{\sqrt{3}}{2}\,i + \dfrac{3}{4}\,i^{2}$

$= \dfrac{1}{4} - \dfrac{3}{4} - \dfrac{\sqrt{3}}{2}\,i$

$\omega^{2} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\,i$ ✓

Matches the polar-form answer exactly — as it must.

(iv) Show that $(\omega + \omega^{2})^{3} = -1$.

Use the result from (ii). From $1 + \omega + \omega^{2} = 0$:

$\omega + \omega^{2} = -1$

$(\omega + \omega^{2})^{3} = (-1)^{3}$

$(\omega + \omega^{2})^{3} = -1$ ✓

A typical exam trick: get the messy expression to equal something simple (like $-1$ or $0$) using $1 + \omega + \omega^{2} = 0$ and $\omega^{3} = 1$.

Try-me 8

Given that $\omega$ is a non-real cube root of unity, evaluate $(1 + \omega)(1 + \omega^{2})$.

Multiply out, then use $\omega + \omega^{2} = -1$ and $\omega^{3} = 1$.

$(1 + \omega)(1 + \omega^{2}) = 1 + \omega^{2} + \omega + \omega^{3}$

$= 1 + (\omega + \omega^{2}) + \omega^{3}$

$= 1 + (-1) + 1$

$= 1$

$1$

Try-me 9

Find the value of $\omega^{10} + \omega^{20}$, where $\omega$ is a non-real cube root of unity.

$\omega^{3} = 1$. Reduce each power mod $3$.

$\omega^{10} = \omega^{9} \cdot \omega = (\omega^{3})^{3} \cdot \omega = 1 \cdot \omega = \omega$

$\omega^{20} = \omega^{18} \cdot \omega^{2} = (\omega^{3})^{6} \cdot \omega^{2} = 1 \cdot \omega^{2} = \omega^{2}$

$\omega^{10} + \omega^{20} = \omega + \omega^{2} = -1$

$\omega^{10} + \omega^{20} = -1$

$-1$

§6

Exam-style combined problems

Exam questions often hide De Moivre's behind some other algebra — solve simultaneous equations or divide complex numbers first, then apply the theorem.

(i) $(2 + 3i)(a + ib) = -1 + 5i$. Express $a + ib$ in polar form, and find $(a + ib)^{11}$.

Method 1 — multiply out, compare real and imaginary.

$(2 + 3i)(a + bi) = -1 + 5i$

$2a + 2bi + 3ai + 3bi^{2} = -1 + 5i$

$(2a - 3b) + i(2b + 3a) = -1 + 5i$

Compare real parts and imaginary parts.

$2a - 3b = -1$

$3a + 2b = 5$

Two equations, two unknowns. Solve them (multiply first by $2$, second by $3$, add).

$a = 1, \quad b = 1 \;\Rightarrow\; a + ib = 1 + i$

Method 2 — divide directly. (Either method is fine — use whichever you prefer.)

$a + bi = \dfrac{-1 + 5i}{2 + 3i} \cdot \dfrac{2 - 3i}{2 - 3i}$

$= \dfrac{(-1 + 5i)(2 - 3i)}{4 + 9} = \dfrac{-2 + 3i + 10i + 15}{13} = \dfrac{13 + 13i}{13} = 1 + i$ ✓

Now polar form of $1 + i$.

$r = \sqrt{2}, \quad \theta = 45^\circ$

$1 + i = \sqrt{2}(\cos 45^\circ + i\sin 45^\circ)$

Apply De Moivre's with $n = 11$.

$(1 + i)^{11} = (\sqrt{2})^{11}\bigl(\cos(11 \cdot 45^\circ) + i\sin(11 \cdot 45^\circ)\bigr)$

$= (\sqrt{2})^{11}(\cos 495^\circ + i\sin 495^\circ)$

$495^\circ - 360^\circ = 135^\circ$.

$= (\sqrt{2})^{11}\left(-\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i\right)$

$= (\sqrt{2})^{11} \cdot \dfrac{\sqrt{2}}{2}(-1 + i)$

Tidy up the surd power: $(\sqrt{2})^{11} \cdot \dfrac{\sqrt{2}}{2} = \dfrac{(\sqrt{2})^{12}}{2} = \dfrac{(2^{1/2})^{12}}{2} = \dfrac{2^{6}}{2} = 2^{5} = 32$.

$(1 + i)^{11} = 32(-1 + i) = -32 + 32i$

(ii) Express $2(1 - i\sqrt{3})$ in polar form, and find $\bigl[2(1 - i\sqrt{3})\bigr]^{3/2}$.

$2(1 - i\sqrt{3}) = 2 - 2\sqrt{3}\,i$

Polar form.

$r = \sqrt{4 + 12} = 4$

$2 - 2\sqrt{3}\,i$ is in the fourth quadrant. $\theta = 300^\circ$ (or $-60^\circ$).

$2 - 2\sqrt{3}\,i = 4(\cos 300^\circ + i\sin 300^\circ)$

Quick reminder

•Whole power ⇒ polar form.

•Fractional power ($\tfrac{3}{2}$) ⇒ general polar form.

$2 - 2\sqrt{3}\,i = 4\bigl(\cos(300 + 360n) + i\sin(300 + 360n)\bigr)$

$(2 - 2\sqrt{3}\,i)^{3/2} = 4^{3/2}\left[\cos\tfrac{3}{2}(300 + 360n) + i\sin\tfrac{3}{2}(300 + 360n)\right]$

$= 8\bigl[\cos(450 + 540n) + i\sin(450 + 540n)\bigr]$

Power $\tfrac{3}{2}$ has denominator $2$ ⇒ two values. Sub $n = 0, 1$.

$n = 0:\ \ 8(\cos 450^\circ + i\sin 450^\circ) = 8(\cos 90^\circ + i\sin 90^\circ) = 0 + 8i$

$n = 1:\ \ 8(\cos 990^\circ + i\sin 990^\circ) = 8(\cos 270^\circ + i\sin 270^\circ) = 0 - 8i$

$\bigl[2(1 - i\sqrt{3})\bigr]^{3/2} = 8i \quad \text{or} \quad -8i$

Try-me 10

Find the two values of $\bigl[2(1 + i\sqrt{3})\bigr]^{3/2}$ in the form $p + qi$.

First quadrant — $\theta = 60^\circ$. Same shape as the worked example, but in the top half.

$2 + 2\sqrt{3}\,i = 4(\cos 60^\circ + i\sin 60^\circ)$

$= 4\bigl(\cos(60 + 360n) + i\sin(60 + 360n)\bigr)$

$(2 + 2\sqrt{3}\,i)^{3/2} = 8\bigl[\cos(90 + 540n) + i\sin(90 + 540n)\bigr]$

$n = 0:\ \ 8(\cos 90^\circ + i\sin 90^\circ) = 8i$

$n = 1:\ \ 8(\cos 630^\circ + i\sin 630^\circ) = 8(\cos 270^\circ + i\sin 270^\circ) = -8i$

$8i \quad \text{or} \quad -8i$

Try-me 11

$(1 + 2i)(a + ib) = -3 + 4i$. Find $a + ib$, express it in polar form, and hence find $(a + ib)^{8}$.

Divide: $a + bi = \dfrac{-3 + 4i}{1 + 2i}$. Multiply top and bottom by $1 - 2i$.

$a + bi = \dfrac{(-3 + 4i)(1 - 2i)}{(1 + 2i)(1 - 2i)} = \dfrac{-3 + 6i + 4i + 8}{1 + 4} = \dfrac{5 + 10i}{5} = 1 + 2i$

$r = \sqrt{1 + 4} = \sqrt{5}, \quad \tan\theta = 2 \;\Rightarrow\; \theta \approx 63.43^\circ$

$(1 + 2i)^{8} = (\sqrt{5})^{8}(\cos 8\theta + i\sin 8\theta)$

$= 625(\cos 507.49^\circ + i\sin 507.49^\circ)$

$= 625(\cos 147.49^\circ + i\sin 147.49^\circ)$

$\approx -527 + 336i$ (exact: $-527 + 336i$ by direct expansion)

$a + ib = 1 + 2i$; $(1 + 2i)^{8} = -527 + 336i$

§7

Proving identities — $\cos n\theta$ in terms of $\cos\theta$

De Moivre's gives us a power tool for proving trig identities. The idea is simple — compute $z^{n}$ two different ways, then equate the parts.

Strategy — four steps

1.Let $z = \cos\theta + i\sin\theta$.

2.Compute $z^{n}$ using De Moivre's: $z^{n} = \cos n\theta + i\sin n\theta$.

3.Compute $z^{n}$ again — this time by binomial expansion of $(\cos\theta + i\sin\theta)^{n}$.

4.Equate the real parts (for $\cos n\theta$ identities) or the imaginary parts (for $\sin n\theta$ identities).

Application 3 — Prove that $\cos 3\theta = 4\cos^{3}\theta - 3\cos\theta$.

Step 1 — let $z = \cos\theta + i\sin\theta$.

Step 2 — De Moivre's with $n = 3$.

$z^{3} = \cos 3\theta + i\sin 3\theta$ (method A)

Step 3 — same $z^{3}$, but binomial expansion of $(\cos\theta + i\sin\theta)^{3}$.

Pascal row $1, 3, 3, 1$ so $(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$.

$(\cos\theta + i\sin\theta)^{3} = \cos^{3}\theta + 3\cos^{2}\theta \cdot i\sin\theta + 3\cos\theta \cdot i^{2}\sin^{2}\theta + i^{3}\sin^{3}\theta$

Now simplify the powers of $i$: $i^{2} = -1$, $i^{3} = -i$.

$= \cos^{3}\theta + 3i\cos^{2}\theta\sin\theta - 3\cos\theta\sin^{2}\theta - i\sin^{3}\theta$

Group real and imaginary.

$z^{3} = \bigl(\cos^{3}\theta - 3\cos\theta\sin^{2}\theta\bigr) + i\bigl(3\cos^{2}\theta\sin\theta - \sin^{3}\theta\bigr)$ (method B)

Step 4 — equate the real parts of (A) and (B).

$\cos 3\theta = \cos^{3}\theta - 3\cos\theta\sin^{2}\theta$

Almost there — but the identity we want has only $\cos\theta$, no $\sin$. Use $\sin^{2}\theta = 1 - \cos^{2}\theta$.

$\cos 3\theta = \cos^{3}\theta - 3\cos\theta(1 - \cos^{2}\theta)$

$= \cos^{3}\theta - 3\cos\theta + 3\cos^{3}\theta$

$\cos 3\theta = 4\cos^{3}\theta - 3\cos\theta$ ✓

Bonus — we also got a sin identity for free. Equating imaginary parts gives $\sin 3\theta = 3\cos^{2}\theta\sin\theta - \sin^{3}\theta$, which simplifies to $\sin 3\theta = 3\sin\theta - 4\sin^{3}\theta$.

Try-me 12

Use De Moivre's and the strategy box to prove that $\sin 3\theta = 3\sin\theta - 4\sin^{3}\theta$.

Same start as Application 3 — but this time equate imaginary parts. At the end use $\cos^{2}\theta = 1 - \sin^{2}\theta$.

$z = \cos\theta + i\sin\theta$

$z^{3} = \cos 3\theta + i\sin 3\theta$

$z^{3} = (\cos\theta + i\sin\theta)^{3}$

$= (\cos^{3}\theta - 3\cos\theta\sin^{2}\theta) + i(3\cos^{2}\theta\sin\theta - \sin^{3}\theta)$

Equate imaginary parts:

$\sin 3\theta = 3\cos^{2}\theta\sin\theta - \sin^{3}\theta$

$= 3(1 - \sin^{2}\theta)\sin\theta - \sin^{3}\theta$

$= 3\sin\theta - 3\sin^{3}\theta - \sin^{3}\theta$

$\sin 3\theta = 3\sin\theta - 4\sin^{3}\theta$ ✓

Equate imaginary parts and use $\cos^{2}\theta = 1 - \sin^{2}\theta$.

Try-me 13

Prove that $\cos 4\theta = 8\cos^{4}\theta - 8\cos^{2}\theta + 1$.

Same method, but expand $(\cos\theta + i\sin\theta)^{4}$. Pascal row $1, 4, 6, 4, 1$. Equate real parts.

$z = \cos\theta + i\sin\theta$, $z^{4} = \cos 4\theta + i\sin 4\theta$

$(\cos\theta + i\sin\theta)^{4} = \cos^{4}\theta + 4i\cos^{3}\theta\sin\theta + 6i^{2}\cos^{2}\theta\sin^{2}\theta + 4i^{3}\cos\theta\sin^{3}\theta + i^{4}\sin^{4}\theta$

$= \cos^{4}\theta + 4i\cos^{3}\theta\sin\theta - 6\cos^{2}\theta\sin^{2}\theta - 4i\cos\theta\sin^{3}\theta + \sin^{4}\theta$

Real part:

$\cos 4\theta = \cos^{4}\theta - 6\cos^{2}\theta\sin^{2}\theta + \sin^{4}\theta$

$= \cos^{4}\theta - 6\cos^{2}\theta(1 - \cos^{2}\theta) + (1 - \cos^{2}\theta)^{2}$

$= \cos^{4}\theta - 6\cos^{2}\theta + 6\cos^{4}\theta + 1 - 2\cos^{2}\theta + \cos^{4}\theta$

$\cos 4\theta = 8\cos^{4}\theta - 8\cos^{2}\theta + 1$ ✓

Expand $(\cos\theta + i\sin\theta)^{4}$, equate real parts, then use $\sin^{2}\theta = 1 - \cos^{2}\theta$.

§8

Reverse direction — $\cos^{n}\theta$ in terms of $\cos n\theta$

In §7 we wrote $\cos 3\theta$ in terms of $\cos\theta$. This section runs the same idea backwards — we'll write $\cos^{3}\theta$ in terms of $\cos 3\theta$ and $\cos\theta$.

The trick uses a beautiful pair of identities. Start from $z = \cos\theta + i\sin\theta$.

$\dfrac{1}{z} = \dfrac{1}{\cos\theta + i\sin\theta}$

Multiply top and bottom by the conjugate $\cos\theta - i\sin\theta$. Denominator becomes $\cos^{2}\theta + \sin^{2}\theta = 1$.

$\dfrac{1}{z} = \cos\theta - i\sin\theta$ (must know this)

Now add and subtract $z$ and $\tfrac{1}{z}$ — the $i\sin\theta$ either cancels or doubles.

$z + \dfrac{1}{z} = 2\cos\theta$

$z - \dfrac{1}{z} = 2i\sin\theta$

Reverse-direction kit — must learn

•$z + \dfrac{1}{z} = 2\cos\theta$

•$z - \dfrac{1}{z} = 2i\sin\theta$

•$z^{n} + \dfrac{1}{z^{n}} = 2\cos n\theta$

•$z^{n} - \dfrac{1}{z^{n}} = 2i\sin n\theta$

The last two follow from De Moivre's — $z^{n} = \cos n\theta + i\sin n\theta$ and $\tfrac{1}{z^{n}} = \cos n\theta - i\sin n\theta$.

Application 4 — Prove that $\cos^{3}\theta = \dfrac{1}{4}\cos 3\theta + \dfrac{3}{4}\cos\theta$.

Start from $z + \tfrac{1}{z} = 2\cos\theta$. Cube both sides.

$(2\cos\theta)^{3} = \left(z + \dfrac{1}{z}\right)^{3}$

$8\cos^{3}\theta = z^{3} + 3z^{2} \cdot \dfrac{1}{z} + 3z \cdot \dfrac{1}{z^{2}} + \dfrac{1}{z^{3}}$

Using $(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$ with $a = z$, $b = \tfrac{1}{z}$.

$8\cos^{3}\theta = z^{3} + 3z + \dfrac{3}{z} + \dfrac{1}{z^{3}}$

Group the $z^{3}$ with $\tfrac{1}{z^{3}}$, and the $3z$ with $\tfrac{3}{z}$.

$8\cos^{3}\theta = \left(z^{3} + \dfrac{1}{z^{3}}\right) + 3\left(z + \dfrac{1}{z}\right)$

Now use the kit identities.

$8\cos^{3}\theta = 2\cos 3\theta + 3(2\cos\theta)$

$8\cos^{3}\theta = 2\cos 3\theta + 6\cos\theta$

Divide both sides by $8$.

$\cos^{3}\theta = \dfrac{1}{4}\cos 3\theta + \dfrac{3}{4}\cos\theta$ ✓

This rewriting is genuinely useful — try integrating $\cos^{3}\theta$ directly versus integrating $\tfrac{1}{4}\cos 3\theta + \tfrac{3}{4}\cos\theta$ and you'll see why.

Try-me 14

Prove that $\sin^{3}\theta = \dfrac{3}{4}\sin\theta - \dfrac{1}{4}\sin 3\theta$.

Start from $z - \dfrac{1}{z} = 2i\sin\theta$, cube it. Watch the $i^{3} = -i$.

$(2i\sin\theta)^{3} = \left(z - \dfrac{1}{z}\right)^{3}$

$8i^{3}\sin^{3}\theta = z^{3} - 3z^{2}\cdot\dfrac{1}{z} + 3z\cdot\dfrac{1}{z^{2}} - \dfrac{1}{z^{3}}$

$-8i\sin^{3}\theta = z^{3} - 3z + \dfrac{3}{z} - \dfrac{1}{z^{3}}$

$-8i\sin^{3}\theta = \left(z^{3} - \dfrac{1}{z^{3}}\right) - 3\left(z - \dfrac{1}{z}\right)$

$-8i\sin^{3}\theta = 2i\sin 3\theta - 3(2i\sin\theta)$

$-8i\sin^{3}\theta = 2i\sin 3\theta - 6i\sin\theta$

Divide both sides by $-8i$:

$\sin^{3}\theta = -\dfrac{1}{4}\sin 3\theta + \dfrac{3}{4}\sin\theta$

$\sin^{3}\theta = \dfrac{3}{4}\sin\theta - \dfrac{1}{4}\sin 3\theta$ ✓

Cube $z - \dfrac{1}{z} = 2i\sin\theta$ and use the kit.

Try-me 15

Prove that $\cos^{4}\theta = \dfrac{1}{8}\cos 4\theta + \dfrac{1}{2}\cos 2\theta + \dfrac{3}{8}$.

Raise $z + \dfrac{1}{z} = 2\cos\theta$ to the fourth power. Pascal row $1, 4, 6, 4, 1$. The middle term $6z^{2}\cdot\dfrac{1}{z^{2}}$ collapses to $6$ — that's where the constant comes from.

$(2\cos\theta)^{4} = \left(z + \dfrac{1}{z}\right)^{4}$

$16\cos^{4}\theta = z^{4} + 4z^{3}\cdot\dfrac{1}{z} + 6z^{2}\cdot\dfrac{1}{z^{2}} + 4z\cdot\dfrac{1}{z^{3}} + \dfrac{1}{z^{4}}$

$16\cos^{4}\theta = z^{4} + 4z^{2} + 6 + \dfrac{4}{z^{2}} + \dfrac{1}{z^{4}}$

$16\cos^{4}\theta = \left(z^{4} + \dfrac{1}{z^{4}}\right) + 4\left(z^{2} + \dfrac{1}{z^{2}}\right) + 6$

$16\cos^{4}\theta = 2\cos 4\theta + 4(2\cos 2\theta) + 6$

$16\cos^{4}\theta = 2\cos 4\theta + 8\cos 2\theta + 6$

Divide by $16$:

$\cos^{4}\theta = \dfrac{1}{8}\cos 4\theta + \dfrac{1}{2}\cos 2\theta + \dfrac{3}{8}$ ✓

Raise to the fourth power and pair up. The middle term gives the constant.

End of lesson

De Moivre's Theorem — powers, roots and identities all from one rule.