De Moivre's theorem and roots
Multiplying complex numbers in trigonometric form means multiplying the moduli and adding the arguments. Doing this again and again gives a quick formula for powers, called De Moivre's theorem. Turned around, it finds roots: for example, all three cube roots of ${8}$, not just the real one.
De Moivre's theorem
Square ${z=r(\cos\theta+i\sin\theta)}$ by multiplying it by itself. The moduli multiply to ${r^2}$, and the arguments add to ${2\theta}$:
${z^2=r^2(\cos 2\theta+i\sin 2\theta)}$
Multiplying by ${z}$ again gives ${z^3=r^3(\cos 3\theta+i\sin 3\theta)}$, and so on. This pattern is named after the French mathematician Abraham de Moivre:
$\begin{align*}&\left[r(\cos\theta+i\sin\theta)\right]^n\\&=r^n(\cos n\theta+i\sin n\theta)\end{align*}$
for every positive whole number ${n}$. In words: raise the modulus to the ${n}$th power, and multiply the argument by ${n}$.
Example 1: Find ${(1+i)^{10}}$.
Solution:
From the last lesson,
$1+i=\sqrt{2}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$
By De Moivre's theorem,
$\begin{align*}&(1+i)^{10}\\&=(\sqrt{2})^{10}\left(\cos\dfrac{10\pi}{4}+i\sin\dfrac{10\pi}{4}\right)\end{align*}$
Now ${(\sqrt{2})^{10}=2^5=32}$. The angle is ${\dfrac{10\pi}{4}=\dfrac{5\pi}{2}}$, which is ${2\pi+\dfrac{\pi}{2}}$. Dropping the full turn,
$\begin{align*}(1+i)^{10}&=32\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)\\&=32i\end{align*}$
Multiplying out ${(1+i)^{10}}$ directly would take many more steps.
Example 2: Find ${(-\sqrt{3}+i)^6}$.
Solution:
From the last lesson,
$\begin{align*}&-\sqrt{3}+i\\&=2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)\end{align*}$
So
$\begin{align*}(-\sqrt{3}+i)^6&=2^6\left(\cos 5\pi+i\sin 5\pi\right)\\&=64(-1+0i)=-64\end{align*}$
Roots of complex numbers
An ${n}$th root of a complex number ${z}$ is a number ${w}$ with ${w^n=z}$. Every nonzero complex number has exactly ${n}$ different ${n}$th roots. For example, ${8}$ has three cube roots, though only one of them, ${2}$, is real.
To find them, use De Moivre's theorem backward: take the ${n}$th root of the modulus, and divide the argument by ${n}$. Since the argument can also be ${\theta+2\pi}$, ${\theta+4\pi}$, and so on, dividing each by ${n}$ gives the other roots.
The ${n}$th roots of ${z=r(\cos\theta+i\sin\theta)}$ are
$\begin{align*}w_k=\sqrt[n]{r}\,\Big(&\cos\dfrac{\theta+2k\pi}{n}\\&+i\sin\dfrac{\theta+2k\pi}{n}\Big)\end{align*}$
for ${k=0,1,2,\ldots,n-1}$
All the roots have the same modulus ${\sqrt[n]{r}}$, so they lie on a circle. Their arguments differ by ${\dfrac{2\pi}{n}}$, so they are evenly spaced around it.
Example 3: Find the three cube roots of ${8}$.
Solution:
Write ${8=8(\cos 0+i\sin 0)}$. Here ${r=8}$, ${\theta=0}$, and ${n=3}$. The modulus of each root is ${\sqrt[3]{8}=2}$, and the arguments are ${\dfrac{0+2k\pi}{3}}$:
- ${k=0}$: ${2(\cos 0+i\sin 0)=2}$
- ${k=1}$: angle ${\dfrac{2\pi}{3}}$, so the root is ${-1+\sqrt{3}\,i}$
- ${k=2}$: angle ${\dfrac{4\pi}{3}}$, so the root is ${-1-\sqrt{3}\,i}$
Check one: Square it first:
$\begin{align*}&(-1+\sqrt{3}\,i)^2\\&=1-2\sqrt{3}\,i-3\\&=-2-2\sqrt{3}\,i\end{align*}$
Multiply by ${-1+\sqrt{3}\,i}$ once more, and you get ${2+6=8}$.
Example 4: Find the four fourth roots of ${-16}$.
Solution:
Write ${-16=16(\cos\pi+i\sin\pi)}$. Each root has modulus ${\sqrt[4]{16}=2}$, and argument ${\dfrac{\pi+2k\pi}{4}}$:
- ${k=0}$: angle ${\dfrac{\pi}{4}}$, so ${\sqrt{2}+\sqrt{2}\,i}$
- ${k=1}$: angle ${\dfrac{3\pi}{4}}$, so ${-\sqrt{2}+\sqrt{2}\,i}$
- ${k=2}$: angle ${\dfrac{5\pi}{4}}$, so ${-\sqrt{2}-\sqrt{2}\,i}$
- ${k=3}$: angle ${\dfrac{7\pi}{4}}$, so ${\sqrt{2}-\sqrt{2}\,i}$
Watch out: Add ${2k\pi}$ to the argument before dividing by ${n}$. Dividing first and then adding ${2k\pi}$ gives the same root over and over.
Solving equations
Finding ${n}$th roots is the same as solving the equation ${z^n=c}$.
Example 5: Solve ${z^6=1}$.
Solution:
The solutions are the sixth roots of ${1=\cos 0+i\sin 0}$. Each has modulus ${1}$ and argument ${\dfrac{2k\pi}{6}=\dfrac{k\pi}{3}}$, for ${k=0}$ to ${5}$:
$z=\cos\dfrac{k\pi}{3}+i\sin\dfrac{k\pi}{3}$
In ${a+bi}$ form, they are ${\pm 1}$ and ${\pm\dfrac{1}{2}\pm\dfrac{\sqrt{3}}{2}i}$ (all four sign choices).
These are called the sixth roots of unity ("unity" means ${1}$). In general, the ${n}$th roots of unity are the corners of a regular polygon with ${n}$ sides on the unit circle.
Summary
- De Moivre's theorem: to find ${z^n}$, raise the modulus to the ${n}$th power and multiply the argument by ${n}$.
- A nonzero complex number has exactly ${n}$ different ${n}$th roots.
- They all have modulus ${\sqrt[n]{r}}$, and arguments ${\dfrac{\theta+2k\pi}{n}}$ for ${k=0}$ to ${n-1}$.
- The roots are evenly spaced on a circle, at the corners of a regular polygon.