De Moivre's theorem

De Moivre's theorem is a powerful tool in mathematics for working with complex numbers. In particular, it helps when dealing with powers and roots of complex numbers. You might apply De Moivre's theorem in a range of disciplines, including AC circuit analysis and signal processing in engineering, computer graphics and navigation, and even in finance to analyse patterns in market data. Use this resource to learn about De Moivre's theorem.

Powers of complex numbers

De Moivre's theorem provides a simple way to raise complex numbers to a power. It states that:\(\DeclareMathOperator{\cis}{cis}\)

If \(z=r\cis(\theta)\), then \(z^{n}=r^{n}\cis(n\theta)\).

Example 1 – using De Moivre's theorem to find the powers of complex numbers

Express \((1-i)^{6}\) in the form \(x+yi\).

First, convert the complex number to its polar form. Go back to Polar form of complex numbers to review how to do this.
\[1-i = \sqrt{2}\cis\left(-\frac{\pi}{4}\right)\]

Now, we can apply De Moivre's theoreom.
\[\begin{align*} \left( \sqrt{2}\cis\left(-\frac{\pi}{4}\right)\right)^{6} & = (\sqrt{2})^{6} \cis \left(6\times-\frac{\pi}{4}\right)\\
& = 8\cis\left(-\frac{3\pi}{2}\right)\\
& = 8i
\end{align*}\]

Exercise – using De Moivre's theorem to find the powers of complex numbers

1. Evaluate the following, giving your answers in polar form, where \(-\pi\leq\theta\leq\pi\).
1. \((\sqrt{3}+i)^{3}\)
2. \((1-i)^{5}\)
3. \((-2\sqrt{3}+2i)^{2}\)
2. Simplify the following, giving your answers in polar form.
1. \((1+i)^{4}(2-2i)^{3}\)
2. \(\dfrac{(2-2\sqrt{3}i)^{4}}{(-1+i)^{6}}\)

Roots of complex numbers

De Moivre's theorem is also helpful for finding all of the possible roots of a complex number. For \(z^{n}=r\cis(\theta)\), there will be \(n\) solutions in the form:

\(z^{\frac{1}{n}} = r^{\frac{1}{n}}\cis \left( \dfrac{\theta+2\pi k}{n} \right)\) where \(k=0,1,\ldots n-1\).

Example – using De Moivre's theorem to find the roots of complex numbers

Solve for the roots of \(z^{n}=1-\sqrt{3}i\).

First, convert the complex number to its polar form.
\[1-\sqrt{3}i = 2\cis\left(-\frac{\pi}{3}\right)\]

Since \(n=4\), this means there are four solutions. Since \(k=0,1,\ldots n-1\), we let \(k=0\), \(k=1\), \(k=2\) and \(k=3\).

For \(k=0\):
\[\begin{align*} z & = 2^{\frac{1}{4}}\cis \left( \frac{-\dfrac{\pi}{3}+2\pi(0)}{4} \right)\\
& = 2^{\frac{1}{4}}\cis\left( -\frac{\pi}{12} \right)
\end{align*}\]

For \(k=1\):
\[\begin{align*} z & = 2^{\frac{1}{4}}\cis \left( \frac{-\dfrac{\pi}{3}+2\pi(1)}{4} \right)\\
& = 2^{\frac{1}{4}}\cis\left( \frac{5\pi}{12} \right)
\end{align*}\]

For \(k=2\):
\[\begin{align*} z & = 2^{\frac{1}{4}}\cis \left( \frac{-\dfrac{\pi}{3}+2\pi(2)}{4} \right)\\
& = 2^{\frac{1}{4}}\cis\left( \frac{11\pi}{12} \right)
\end{align*}\]

For \(k=3\):
\[\begin{align*} z & = 2^{\frac{1}{4}}\cis \left( \frac{-\dfrac{\pi}{3}+2\pi(3)}{4} \right)\\
& = 2^{\frac{1}{4}}\cis\left( \frac{17\pi}{12} \right)
\end{align*}\]

The four solutions for \(z\) are \(2^{\frac{1}{4}}\cis\left( -\dfrac{\pi}{12} \right)\), \(2^{\frac{1}{4}}\cis\left( \dfrac{5\pi}{12} \right)\), \(2^{\frac{1}{4}}\cis\left( \dfrac{11\pi}{12} \right) \) and \(2^{\frac{1}{4}}\cis\left( \dfrac{17\pi}{12} \right)\).

This can be represented graphically:

Exercise – using De Moivre's theorem to find the roots of complex numbers

1. Solve the following, giving your answers in polar form, where \(-\pi\leq\theta\leq\pi\).
1. \(z^{3}=-1\)
2. \(z^{4}=16i\)
3. \(z^{3}=\sqrt{6}-\sqrt{2}i\)
2. If \(\sqrt{3}+i\) is one solution of \(z^{3}=8i\), use a diagram to find the other solutions in rectangular form.

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