12. Taking powers of complex numbers

← 11. Multiplication and de Moivre's Theorem

Now we know how easy it is to multiply together two complex numbers when they are written in modulus-argument form. How about squaring a complex number. This is just a special case of multiplication, where the number is being multiplied by itself. Let's start with complex numbers of modulus 1, and adapt to the more general case later. Take the above expression for $zw$ and make $w$ be the same as $z$. ie. if $z=w=\cos{\theta}+i\sin{\theta}$, you get:

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

How about $z^3$? Well, you just multiply $z^2$ by $z$, which yields:

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

In fact, repeated multiplications like this give us a general statement and the name of it is De Moivre's Theorem:

Theorem 12.1

De Moivre's Theorem

For any integer $n$ and real number $\theta$,

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

From the picture of multiplication of complex numbers as rotations, de Moivre's theorem should become relatively intuitive. Note that De Moivre's Theorem is for complex numbers with modulus 1. However, the theorem can be extended to include any complex number $z$ with $|z|=r$ i.e. $z=r(\cos{\theta}+i\sin{\theta})$:

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

Let's look at the proof of De Moivre's Theorem.

Proof

Proof of theorem 12.1

We use mathematical induction to prove the theorem.

Case 1 : $n \geq 0$

For any integer $n\geq 0$ let $P(n)$ be the proposition that
$$(\cos{\theta}+i\sin{\theta})^n=\cos{(n\theta)}+i\sin{(n\theta)}$$

Base case: The statement $P(0)$ says that

$$(\cos{\theta}+i\sin{\theta})^0=\cos{(0)}+i\sin{(0)}=1$$ which is true.

Induction Hypothesis (IH): We now assume that $P(m)$ is true for some non-negative integer $m$ $$(\cos{\theta}+i\sin{\theta})^m=\cos{(m\theta)}+i\sin{(m\theta)}$$

Inductive Step: It now remains for us to show that $P(m+1)$ holds true

$$\begin{aligned}\cos{((m+1)\theta)}+i\sin{((m+1)\theta)}&=(\cos{\theta}+i\sin{\theta})^{m+1}\\ &=(\cos{\theta}+i\sin{\theta})^m(\cos\theta +i\sin\theta)\\ &=(\cos{(m\theta)}+i\sin{(m\theta)})(\cos\theta +i\sin\theta) && \textbf{IH}\\ &=\cos{((m+1)\theta)}+i\sin{((m+1)\theta)} && \textbf{Theorem \href{../sec11-de-moivre-theorem/#thm-11-1}{11.1}}\\\end{aligned}$$

So $P(m+1)$ holds true. By the principle of mathematical induction $P(n)$ is true for all non-negative integers $n$.

Proof

Case 2: $n <0 $

Let $n=-k$ where $n$ is any negative integer, so $k$ is a positive integer.

$$\begin{aligned}(\cos\theta +i\sin\theta)^n&=(\cos\theta +i\sin\theta)^n\\ &=(\cos\theta +i\sin\theta)^{-k}\\ &=\dfrac1{(\cos\theta +i\sin\theta)^k}\\ &=\dfrac1{\cos{(k\theta)} +i\sin{(k\theta})} && \textbf{Case 1}\\ &=\dfrac1{\cos{(k\theta)} +i\sin{(k\theta})} \times \dfrac{{\cos{(k\theta)} -i\sin{(k\theta})}}{{\cos{(k\theta)} -i\sin{(k\theta})}}\\ &=\dfrac{\cos{(k\theta)} -i\sin{(k\theta)}}{\cos^2{(k\theta)} -i^2\sin^2{(k\theta)}}\\ &=\dfrac{\cos{(k\theta)} -i\sin{(k\theta)}}{\cos^2{(k\theta)} +\sin^2{(k\theta)}}\\ &=\cos{(k\theta)} -i\sin{(k\theta)}\\ &=\cos{(-n\theta)} -i\sin{(-n\theta)}\\ &=\cos{(n\theta)} +i\sin{(n\theta)} \\\end{aligned}$$

We have now proved that $(\cos\theta +i\sin\theta)^n=\cos{(n\theta)} +i\sin{(n\theta)}$ for any integer $n$. $\Box $

Worked examples 12.1
  1. Use De Moivre's Theorem to write $(1+i)^{20}$ in a cartesian from.
  2. Write $(\sqrt{3}+i)^3(1+\sqrt{3}i)^5$ in a modulus-argument form.
  3. Write $\dfrac{(1+i)^{10}}{1+\sqrt{3}i}$ in a modulus-argument form.
Show solutions

  1. We first need to write this number in modulus argument as that is how de Moivre's theorem is stated. We can write:

    $$(1+i)=\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$$

    and therefore:

    $$\begin{aligned}(1+i)^{20}& = (\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}))^{20}=\sqrt{2}^{20}(\cos\frac{20\pi}{4}+i\sin\frac{20\pi}{4})\\ & = 2^{10}(\cos 5\pi+i\sin 5\pi)\end{aligned}$$

    But we know that we can always subtract any integer multiple of $2\pi$ from an argument and it will still be the same complex number. Because the angle here is $5\pi$ if we subtract 2 lots of $2\pi$ we will end up with $\pi$ which is a principal argument, so:

    $$(1+i)^{20}=2^{10}(\cos \pi+i\sin \pi)=2^{10}(-1+0 i)=-2^{10}$$

  2. We will do the same thing for the next example, but without the explanatory steps. Go through and make sure that you can see what is happening here:

    $$\begin{aligned}(\sqrt{3}+i)^3(1+\sqrt{3}i)^5& = (2(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}))^3((2(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}))^5\\ & = 2^3(\cos\frac{3\pi}{6}+i\sin\frac{3\pi}{6})2^5(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3})\\ & = 2^32^5(\cos(\frac{3\pi}{6}+\frac{10\pi}{6})+i\sin(\frac{3\pi}{6}+\frac{10\pi}{6})\\ & = 2^8(\cos\frac{13\pi}{6}+i\sin\frac{13\pi}{6})\\ & = 2^8(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})\\\end{aligned}$$

  3. And again, make sure that you can follow all steps:

    $$\begin{aligned}\frac{(1+i)^{10}}{1+\sqrt{3}i}& = (1+i)^{10}(1+\sqrt{3}i)^{-1}\\ & = (\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}))^{10} (2(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}))^{-1}\\ & = \frac{2^5}{2}(\cos\frac{10\pi}{4}+i\sin\frac{10\pi}{4})(\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3}))\\ & = 2^4(\cos(\frac{10\pi}{4}-\frac{\pi}{3})+i\cos(\frac{10\pi}{4}-\frac{\pi}{3}))\\ & = 2^4(\cos(\frac{13\pi}{6})+i\cos(\frac{13\pi}{6}))\\ & = 2^4(\cos(\frac{\pi}{6})+i\cos(\frac{\pi}{6}))\end{aligned}$$
Check your understanding 12.1

  1. Write the following complex numbers in a cartesian form:

    1. $(1-\sqrt{3}i)^5$
    2. $(2\sqrt{3}+2i)^5$
    3. $(1-i)^8$
    4. $(1-\sqrt{3}i)^5(\sqrt{3}+i)^2$
    5. $\dfrac{(1-\sqrt{3}i)^5}{(1-i)^6}$
  2. By considering the expansion of $(1+i)^{10}$, show that $\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}=32$

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

    1. Show that $2\cos{\theta}=z+\dfrac1{z}$.
    2. Show that $2\cos{(n\theta)}=z^n+\dfrac1{z^n}$.
  4. Find the exact value of $\dfrac1{(\sqrt{3}+i)^6}$, clearly showing your working.
    Show answer

    -1/64

    1. Express $\dfrac{\sqrt{3}}{2}-\dfrac1{2}i$ in the modulus-argument form.
      Show answer

      MOD 1 ARG = -PI/6

    2. Hence show that $(\dfrac{\sqrt{3}}{2}-\dfrac1{2}i)^9=ci$ where $c$ is a real number to be found.
      Show answer

      C=1

    3. Find one pair of possible values of positive integers $m$ and $n$ such that $$\Big(\dfrac{\sqrt{3}}{2}-\dfrac1{2}i\Big)^m=\Big(\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i\Big)^n$$
      Show answer

      m=6 and n=4

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