Question

Find the trigonometric form of the number 3 + √3i.

Answer 1

Any complex number $z$ can be written in trigonometric form as

$z = |z| e^{i\arg(z)} = |z| \left(\cos(\arg(z)) + i \sin(\arg(z))\right)$

where $|z|$ is the modulus of $z$ and $\arg(z)$ is its argument, i.e. the angle $z$ makes with the positive real axis in the complex plane.

We have

$|z| = \sqrt{3^2 + \left(\sqrt3\right)^2} = \sqrt{12} = 2\sqrt3$

and

$\arg(z) = \tan^{-1}\left(\dfrac{\sqrt3}3\right) = \tan^{-1}\left(\dfrac1{\sqrt3}\right) = \dfrac\pi6$

Then

$3 + \sqrt3 \, i = \boxed{2\sqrt 3 \left(\cos\left(\dfrac\pi6\right) + i \sin\left(\dfrac\pi6\right)\right)}$