# Trigonometric Form of Complex Numbers

## Key Questions

• DeMoivre's Theorem expand's on Euler's formula:
${e}^{i x} = \cos x + i \sin x$

DeMoivre's Theorem says that:

• ${\left({e}^{i x}\right)}^{n} = {\left(\cos x + i \sin x\right)}^{n}$
• ${\left({e}^{i x}\right)}^{n} = {e}^{i n x}$
• ${e}^{i n x} = \cos \left(n x\right) + i \sin \left(n x\right)$
• $\cos \left(n x\right) + i \sin \left(n x\right) \equiv {\left(\cos x + i \sin x\right)}^{n}$

Example:
$\cos \left(2 x\right) + i \sin \left(2 x\right) \equiv {\left(\cos x + i \sin x\right)}^{2}$

${\left(\cos x + i \sin x\right)}^{2} = {\cos}^{2} x + 2 i \cos x \sin x + {i}^{2} {\sin}^{2} x$

However, ${i}^{2} = - 1$

${\left(\cos x + i \sin x\right)}^{2} = {\cos}^{2} x + 2 i \cos x \sin x - {\sin}^{2} x$

Resolving for real and imaginary parts of $x$:
${\cos}^{2} x - {\sin}^{2} x + i \left(2 \cos x \sin x\right)$

Comparing to $\cos \left(2 x\right) + i \sin \left(2 x\right)$

$\cos \left(2 x\right) = {\cos}^{2} x - {\sin}^{2} x$
$\sin \left(2 x\right) = 2 \sin x \cos x$
These are the double angle formulae for $\cos$ and $\sin$

This allows us to expand $\cos \left(n x\right)$ or $\sin \left(n x\right)$ in terms of powers of $\sin x$ and $\cos x$

DeMoivre's theorem can be taken further:
Given $z = \cos x + i \sin x$

${z}^{n} = \cos \left(n x\right) + i \sin \left(n x\right)$

${z}^{- n} = {\left(\cos x + i \sin x\right)}^{- n} = \frac{1}{\cos \left(n x\right) + i \sin \left(n x\right)}$

${z}^{- n} = \frac{1}{\cos \left(n x\right) + i \sin \left(n x\right)} \times \frac{\cos \left(n x\right) - i \sin \left(n x\right)}{\cos \left(n x\right) - i \sin \left(n x\right)} = \frac{\cos \left(n x\right) - i \sin \left(n x\right)}{{\cos}^{2} \left(n x\right) + {\sin}^{2} \left(n x\right)} = \cos \left(n x\right) - i \sin \left(n x\right)$

${z}^{n} + {z}^{- n} = 2 \cos \left(n x\right)$
${z}^{n} - {z}^{- n} = 2 i \sin \left(n x\right)$

So, if you wanted to express ${\sin}^{n} x$ in terms of multiple angles of $\sin x$ and $\cos x$:
${\left(2 i \sin x\right)}^{n} = {\left(z - \frac{1}{z}\right)}^{n}$

Expand and simply, then input values for ${z}^{n} + {z}^{- n}$ and ${z}^{n} - {z}^{- n}$ where necessary.

However, if it involved ${\cos}^{n} x$, then you would do ${\left(2 \cos x\right)}^{n} = {\left(z + \frac{1}{z}\right)}^{n}$ and follow the similar steps.

• Depending on what you need to do with your complex numbers, the trigonometric form can be very useful or very thorny.

For example, let ${z}_{1} = 1 + i$, ${z}_{2} = \sqrt{3} + i$ and ${z}_{3} = - 1 + i \sqrt{3}$.
Let's compute the two trigonometric forms:
${\theta}_{1} = \arctan \left(1\right) = \frac{\pi}{4}$ and ${\rho}_{1} = \sqrt{1 + 1} = \sqrt{2}$
${\theta}_{2} = \arctan \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ and ${\rho}_{2} = \sqrt{3 + 1} = 2$
${\theta}_{3} = \pi + \arctan \left(- \sqrt{3}\right) = \frac{2}{3} \pi$ and ${\rho}_{3} = \sqrt{1 + 3} = 2$
So the trigonometric forms are:
${z}_{1} = \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$
${z}_{2} = 2 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right)$
${z}_{3} = 2 \left(\cos \left(\frac{2}{3} \pi\right) + i \sin \left(\frac{2}{3} \pi\right)\right)$

Let's say you want to compute ${z}_{1} + {z}_{2} + {z}_{3}$. If you use the algebraic form, you get
${z}_{1} + {z}_{2} + {z}_{3} = \left(1 + i\right) + \left(\sqrt{3} + i\right) + \left(- 1 + i \sqrt{3}\right) = \sqrt{3} + i \left(2 + \sqrt{3}\right)$
Quite easy. Now try with the trigonometric form...
${z}_{1} + {z}_{2} + {z}_{3} = \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right) + 2 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right) + 2 \left(\cos \left(\frac{2}{3} \pi\right) + i \sin \left(\frac{2}{3} \pi\right)\right)$
it turns out that the shortest way to add these two expressions is to solve cosines and sines, which means... turning to the algebraic form!
The algebraic form is often the best form to choose in adding complex numbers.

Multiplication
Now we try to compute ${z}_{1} \cdot {z}_{2} \cdot {z}_{3}$. Using algebraic forms requires a lot of annoying computations. But solving this product with the trigonometric forms is simpler:
${z}_{1} \cdot {z}_{2} \cdot {z}_{3} = \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right) \cdot 2 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right) \cdot 2 \left(\cos \left(\frac{2}{3} \pi\right) + i \sin \left(\frac{2}{3} \pi\right)\right) = 4 \sqrt{2} \left(\cos \left(\frac{\pi}{4} + \frac{\pi}{6} + \frac{2}{3} \pi\right) + i \sin \left(\frac{\pi}{4} + \frac{\pi}{6} + \frac{2}{3} \pi\right)\right) = 4 \sqrt{2} \left(\cos \left(\frac{13}{12} \pi\right) + i \sin \left(\frac{13}{12} \pi\right)\right)$
The ingredients to prove that the second equality holds come from trigonometry: the two addition formulae
$\sin \left(\alpha + \beta\right) = \sin \left(\alpha\right) \cos \left(\beta\right) + \sin \left(\beta\right) \cos \left(\alpha\right)$
$\cos \left(\alpha + \beta\right) = \cos \left(\alpha\right) \cos \left(\beta\right) - \sin \left(\alpha\right) \sin \left(\beta\right)$
Multiplication of complex numbers is even cleaner (but conceptually not easier) in exponential form.

In some sense, the trigonometric form is a sort of in-between form between the algebraic and the exponential forms. The trigonometric form is the way to switch between these two. In this sense it's a sort of a "dictionary" to "translate" forms.

• Let $z = x + i y$ a complex number in algebraic form.

$z = r \left(\cos \phi + i \sin \phi\right)$ is its trigonometric form, where:

$r = \sqrt{{x}^{2} + {y}^{2}}$ is the modulus of the number and

• if $x > 0$

$\phi = \arctan \left(\frac{y}{x}\right)$ ,

• if $x < 0$

$\phi = \arctan \left(\frac{y}{x}\right) + \pi$,

• if $x = 0$ and $y > 0$

$\phi = \frac{\pi}{2}$,

• if $x = 0$ and $y < 0$

$\phi = \frac{3}{2} \pi$

• if $x = y = 0$

It's all zero!