Question

Express the complex number in trigonometric form.

-6 + 6 $\sqrt{3}$ i

Answer 1

Answer:

$\large\boxed{-6+6\sqrt3i=12\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)}$

Step-by-step explanation:

Look at the picture.

The trigonometric form of a complex number:

$z=|z|(\cos\alpha+i\sin\alpha)$

where:

$|z|=\sqrt{a^2+b^2}\\\\\cos\alpha=\dfrac{a}{|z|}\\\\\sin\alpha=\dfrac{b}{|z|}$

We have the complex number:

$z=-6+6\sqrt3i\to a=-6,\ b=6\sqrt6$

Substitute:

$|z|=\sqrt{(-6)^2+(6\sqrt3)^2}=\sqrt{36+108}=\sqrt{144}=12$

$\cos\alpha=\dfrac{-6}{12}=-\dfrac{1}{2}\\\\\sin\alpha=\dfrac{6\sqrt3}{12}=\dfrac{\sqrt3}{2}$

Therefore

$\alpha=\dfrac{2\pi}{3}$

Finally:

$-6+6\sqrt3i=12\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)$