Question

Express the complex number in trigonometric form.

-6 + 6 
`√(3)` i

Answer 1

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

Explanation:

Look at the picture.

The trigonometric form of a complex number:

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

where:

`|z|=√(a^2+b^2)\\\\\cos\alpha=(a)/(|z|)\\\\\sin\alpha=(b)/(|z|)`

We have the complex number:

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

Substitute:

`|z|=√((-6)^2+(6\sqrt3)^2)=√(36+108)=√(144)=12`

`\cos\alpha=(-6)/(12)=-(1)/(2)\\\\\sin\alpha=(6\sqrt3)/(12)=(\sqrt3)/(2)`

Therefore

`\alpha=(2\pi)/(3)`

Finally:

`-6+6\sqrt3i=12\left(\cos(2\pi)/(3)+i\sin(2\pi)/(3)\right)`