Question

What is the trigonometric form of `(-2-9i)`?

Answer 1

`sqrt(85)e^(iarccos(-2/85))`

Explanation:

You first need the module of this complex number, given by the formula `abs(-2-9i) = sqrt((-9)^2 + (-2)^2) = sqrt85`.

You can now factorize the complex number by its module : `-2-9i = sqrt(85)(-2/85 -i9/85)`. You can now say that `EEtheta in RR` such that `cos(theta) = -2/85` and `sin(theta) = -9/85`.

So `theta = arccos(-2/85)`.