De Moivre's Theorem

De Moivre's theorem gives a formula for computing powers of complex numbers. We first gain some intuition for de Moivre's theorem by considering what happens when we multiply a complex number by itself.

Recall that using the polar form, any complex number $z=a+ib$ can be represented as $z = r ( \cos \theta + i \sin \theta )$ with

\[\begin{array}{rl} \mbox{Absolute value: } & r = \sqrt{ a^2 + b^2 } \\
\mbox{Argument } \theta \text{ subject to: } & \cos{\theta} = \frac{a}{r},\ \sin{\theta}=\frac{b}{r}. \end{array}\]

Then squaring the complex number $z$ gives

$$\begin{aligned} z^2 &= \big( r ( \cos \theta + i \sin \theta )\big) ^2\\ &= r^2 \left( \cos \theta + i \sin \theta \right)^2\\ &= r^2 \left( \cos \theta \cos \theta + i \sin \theta \cos \theta + i \sin \theta \cos \theta + i^2 \sin \theta \sin \theta \right) \\ &= r^2 \big( ( \cos \theta \cos \theta - \sin \theta \sin \theta ) + i ( \sin \theta \cos \theta + \sin \theta \cos \theta )\big) \\ &= r^2 \left( \cos 2\theta + i \sin 2\theta \right). \end{aligned}$$

This shows that by squaring a complex number, the absolute value is squared and the argument is multiplied by $2$. For $n \geq 3$, de Moivre's theorem generalizes this to show that to raise a complex number to the $n^\text{th}$ power, the absolute value is raised to the $n^\text{th}$ power and the argument is multiplied by $n$.

De Moivre's Theorem:

For any complex number $x$ and any integer $n$,

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

We'll prove by induction.

We have

$$\begin{aligned} z^{n} &= \big(r(\cos(\theta) + i\sin(\theta)\big)^{n}\\ &= r^{n}\big(\cos(\theta) + i\sin(\theta)\big)^{n}. \end{aligned}$$

Let's focus on the second part: $\big(\cos(\theta) + i\sin(\theta)\big)^{n}$. For $n = 1$, we have

$$\big(\cos(\theta) + i\sin(\theta)\big)^{1} = \cos(1\cdot \theta) + i\sin(1\cdot \theta),$$

which is true.

We can assume the same formula is true for $n = k$, so we have

$$\big(\cos(\theta) + i\sin(\theta)\big)^{k} = \cos(k\theta) + i\sin(k\theta).$$

For $n = k + 1$, we expect to have

$$\big(\cos(\theta) + i\sin(\theta)\big)^{k + 1} = \cos\big((k + 1)\theta\big) + i\sin\big((k + 1)\theta\big).$$

We get

$$\begin{aligned} \big(\cos(\theta) + i\sin(\theta)\big)^{k + 1} &= \big(\cos(\theta) + i\sin(\theta)\big)^{k}\big(\cos(\theta) + i\sin(\theta)\big)^{1}\\ &= \big(\cos(k\theta) + i\sin(k\theta)\big)\big(\cos(1\cdot \theta) + i\sin(1\cdot \theta)\big) && (\text{We assume this to be true for } x = k.)\\ &= \cos(k\theta)\cos(\theta) + \cos(k\theta)i\sin(\theta) + i\sin(k\theta)\cos(\theta) + i^{2}\sin(k\theta)\sin(\theta) && (\text{We have } i^{2} = -1.)\\ &= \cos(k\theta)\cos(\theta) - \sin(k\theta)\sin(\theta) + i\big(\cos(k\theta)\sin(\theta) + \sin(k\theta)\cos(\theta)\big)\\ &= \cos(k\theta + \theta) + i\sin(k\theta + \theta) && (\text{deducted from the trigonometry rules})\\ &= \cos\big((k+1)\theta\big) + i\sin\big((k+1)\theta\big). \end{aligned}$$

Thus, for $n = k + 1,$ we have $\big(\cos(\theta) + i\sin(\theta)\big)^{k + 1} = \cos\big((k+1)\theta\big) + i\sin\big((k+1)\theta\big),$ as expected.

As the theorem is true for $n = 1$ and $n = k + 1$, it is true for all $n \geq 1$. $_\square$

Note that in de Moivre's theorem, the complex number is in the form $z = r ( \cos \theta + i \sin \theta ) .$ For complex numbers in the general form $z = a + bi$, it may be necessary to first compute the absolute value and argument to convert $z$ to the form $r ( \cos \theta + i \sin \theta )$ before applying de Moivre's theorem.

Evaluate $( 1 - i )^{6}$.

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In order to express $z = 1 - i$ in the form $r (\cos \theta + i \sin \theta),$ we calculate the absolute value $r$ and argument $\theta$ as follows:

\[\begin{align} \mbox{Absolute value}: & r = \sqrt{ 1^2 + (-1) ^2 } = \sqrt{2} \\
\mbox{Argument}: & \theta = \arctan \frac{-1 }{1} = -\frac{\pi}{4}. \end{align}\]

Now, applying DeMoivre's theorem, we obtain

$$\begin{aligned} z^{6} &= \left[ \sqrt{2} \left( \cos \left( -\frac{ \pi}{4} \right) + i \sin \left( -\frac{\pi}{4} \right) \right) \right]^{6} \\ &= \sqrt{2}^{6} \left[ \cos \left(- \frac{ 6\pi } { 4} \right) + i \sin \left(- \frac{6\pi}{4}\right) \right] \\ &= 2^3 \left[ \cos \left(- \frac{ 3\pi }{2} \right) + i \sin \left( - \frac{3\pi}{2} \right) \right] \\ &= 8 ( 0 + 1 i ) \\ &= 8i.\ _\square \end{aligned}$$

Evaluate $\left( \frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2} i \right)^{1000} .$

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In order to express $z = \left( \frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2} i \right)$ in the form $r (\cos \theta + i \sin \theta),$ we calculate the absolute value $r$ and argument $\theta$ as follows:

\[\begin{align} \mbox{Absolute value}: & r = \sqrt{ \left( \frac{\sqrt{2}}{2}\right)^2 + \left( \frac{\sqrt{2}}{2}\right)^2 } = 1 \\
\mbox{Argument}: & \theta = \arctan 1 = \frac{\pi}{4}. \end{align}\]

Now, applying DeMoivre's theorem, we obtain

$$\begin{aligned} z^{1000} &= \Bigg( \cos \left( \frac{ \pi}{4} \right) + i \sin \left( \frac{\pi}{4} \right) \Bigg)^{1000} \\ &= \cos \left( \frac{ 1000\pi }{ 4} \right) + i \sin \left( \frac{1000\pi}{4} \right) \\ &= \cos 250\pi + i \sin 250 \pi \\ &= \cos (0 + 125 \times 2\pi) + i \sin (0 + 125 \times 2\pi)\\ & = 1.\ _\square \end{aligned}$$

Evaluate $\big( 1 + \sqrt{3} i \big)^{2013}.$

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In order to express $z = 1 + \sqrt{3} i$ in the form $r (\cos \theta + i \sin \theta),$ we calculate the absolute value $r$ and argument $\theta$ as follows:

\[\begin{align} \mbox{Absolute value}: & r = \sqrt{ 1^2 + \big(\sqrt{3}\big) ^2 } = \sqrt{4} = 2 \\
\mbox{Argument}: & \theta = \arctan \frac{\sqrt{3} } {1} = \frac{\pi}{3}. \end{align}\]

Now, applying DeMoivre's theorem, we obtain

$$\begin{aligned} z^{2013} &= \Bigg( 2 \left( \cos \frac{ \pi}{3} + i \sin \frac{\pi}{3} \right) \Bigg)^{2013} \\ &= 2^{2013} \left( \cos \frac{ 2013 \pi } { 3} + i \sin \frac{2013\pi}{3} \right) \\ &= 2^{2013} ( - 1 + 0 i ) \\&= - 2^{2013}.\ _\square \end{aligned}$$

De Moivre's Theorem:

For any complex number $x$ and any integer $n$,

$$( \cos x + i \sin x )^n = \cos ( nx) + i \sin (nx).$$

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Proof: We prove this formula by induction on $n$ and by applying the trigonometric sum and product formulas. We first consider the non-negative integers. The base case $n=0$ is clearly true. For the induction step, observe that

$$\begin{array} { l l } ( \cos x + i \sin x)^{k+1} & = (\cos x + i \sin x )^k \times ( \cos x + i \sin x ) \\ & = \big( \cos (kx) + i \sin (kx) \big) ( \cos x + i \sin x ) \\ & = \cos (kx) \cos x - \sin(kx) \sin x + i\big( \sin (kx) \cos x + \cos(kx) \sin x\big) \\ & = \cos \big[(k+1)x\big] + i \sin \big[(k+1)x\big].\ _\square \end{array}$$

Note that the proof above is only valid for integers $n$. There is a more general version, in which $n$ is allowed to be a complex number. In this case, the left-hand side is a multi-valued function, and the right-hand side is one of its possible values.

Euler's formula for complex numbers states that if $z$ is a complex number with absolute value $r_z$ and argument $\theta_z$, then

$$z = r_z e^{i \theta_z}.$$

The proof of this is best approached using the (Maclaurin) power series expansion and is left to the interested reader. With this, we have another proof of De Moivre's theorem that directly follows from the multiplication of complex numbers in polar form.

Show that $\cos (5\theta) = cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta.$

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Applying De Moivre's theorem for $n= 5$, we have

$$\cos (5 \theta) + i \sin ( 5 \theta) = ( \cos \theta + i \sin \theta) ^ 5 .$$

Expand the RHS using the binomial theorem and compare real parts to obtain

$$\cos ( 5 \theta) = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta.\ _\square$$

Note: For an integer $n$, we can express $\cos ( n \theta)$ solely in terms of $\cos \theta$ by using the identity $\sin^2 \theta = 1 - \cos^2 \theta$. This is known as the Chebyshev polynomial of the first kind.

Evaluate $\sin (0\theta) + \sin (1 \theta) + \sin (2 \theta) + \cdots + \sin (n \theta).$

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Applying De Moivre's formula, this is equivalent to the imaginary part of

$$( \cos \theta + i \sin \theta)^0 + ( \cos \theta + i \sin \theta)^1 + ( \cos \theta + i \sin \theta) ^2 + \cdots + ( \cos \theta + i \sin \theta)^n.$$

Interpreting this as a geometric progression, the sum is

$$\frac{ (\cos \theta + i \sin \theta)^{n+1} -1} {( \cos \theta + i \sin \theta) - 1 }$$

as long as the ratio is not 1, which means $\theta \neq 2k \pi$. $\big($Note that in this case, we get that each term $\sin (k\theta)$ is 0, and hence the sum is 0.$\big)$

Converting this to polar form, we obtain

$$\frac{ e^{i (n+1) \theta} - 1 } { e^{i\theta} -1 } = \frac{ e^{ i \left( \frac{n+1}{2} \right)\theta} } {e^{i \frac{1}{2} \theta} } \times \frac{e^{ i \left( \frac{n+1}{2} \right)\theta} - e^{ - i \left( \frac{n+1}{2} \right)\theta} } { e^{ i \frac{1}{2} \theta} - e^{-i \frac{1}{2} \theta} } = e^{ i\frac{n}{2} \theta} \frac{2i \sin \left[ ( \frac{n+1}{2})\theta \right] } { 2i \sin \left(\frac{1}{2} \theta\right)} .$$

Taking imaginary parts, we obtain

$$\frac{ \sin \left( \frac{n}{2} \theta \right) \sin \left( \frac{n+1}{2} \theta \right) } { \sin \left( \frac{1}{2} \theta \right) }.\ _\square$$

The $n^\text{th}$ roots of unity are the complex solutions to the equation

$$z^n = 1.$$

Suppose complex number $z = a + bi$ is a solution to this equation, and consider the polar representation $z = r e^{i\theta}$, where $r = \sqrt{a^2 + b^2}$ and $\tan \theta = \frac{b}{a}, 0 \leq \theta < 2\pi$. Then, by De Moivre's theorem, we have

$$1 = z^n = \big(r e^{i\theta} \big) ^n = r^n (\cos \theta + i \sin \theta)^n = r^n (\cos n \theta + i \sin n \theta).$$

This implies $r^n = 1$ and, since $r$ is a real, non-negative number, we have $r = 1.$ Also, $n \theta = 2k \pi$ or $\theta = \frac{2k \pi}{n}$ for some integer $k$. Now, the values $k = 0, 1, 2, \ldots, n-1$ give distinct values of $\theta$ and, for any other value of $k$, we can add or subtract an integer multiple of $n$ to reduce to one of these values of $\theta$.

Therefore, the $n^\text{th}$ roots of unity are the complex numbers

$$e^{\frac{2k\pi }{ n} i} = \cos \left( \frac{2k\pi }{ n } \right) + i \sin \left( \frac{2k\pi }{ n } \right) \text{ for } k = 0, 1, 2, \ldots, n-1.$$

Observe that this gives $n$ complex $n^\text{th}$ roots of unity, as we know from the fundamental theorem of algebra. Since all of the complex roots of unity have absolute value 1, these points all lie on the unit circle. Furthermore, since the angle between any two consecutive roots is $\frac{2\pi}{n}$, the complex roots of unity are evenly spaced around the unit circle.

What are the complex solutions to the equation $z = \sqrt[3]{1}?$

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Cubing both sides gives $z^3 = 1,$ implying $z$ is a $3^\text{rd}$ root of unity. By the above, the $3^\text{rd}$ roots of unity are

$$e^{ \frac{2k\pi }{ 3 } i} = \cos \left( \frac{2k\pi }{ 3} \right) + i \sin \left( \frac{2k\pi }{ 3 } \right) \text{ for } k = 0,1,2.$$

This gives the roots of unity $1, e^{\frac{2\pi}{3} i}, e^{\frac{4\pi}{3} i}$, or

$$1,\quad -\frac{1}{2} + \frac{\sqrt{3}}{2} i,\quad -\frac{1}{2} - \frac{ \sqrt{3}}{2}i.\ _\square$$

Note: Another way to solve this equation would be to factorize $z^3 -1 = (z-1) (z^2 + z + 1)$. Then the solutions are $z=1$ and the solutions to the quadratic equation $z^2 + z + 1=0$, which can be found using the quadratic formula.

Given positive integer $n$, let $\zeta = e^{\frac{2k\pi }{ n} i }$ for some $k = 1, 2, \ldots, n-1$, i.e., $\zeta$ is one of the $n^\text{th}$ root of unity that is not equal to $1$. Show that

$$1 + \zeta + \zeta^2 + \cdots + \zeta^{n-1} = 0.$$

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Since $\zeta$ is an $n^\text{th}$ root of unity, we have $\zeta^n = 1$. Then

$$0 = 1 - \zeta^n = (1- \zeta)\big( 1 + \zeta + \zeta^2 + \cdots + \zeta^{n-1}\big).$$

Since $\zeta \ne 1$, we have $1 + \zeta + \zeta^2 + \cdots + \zeta^{n-1} = 0.\ _\square$