Question

What is the trigonometric form of `(12-2i)`?

Answer 1

`2sqrt37 [ cos(-0.165) + isin(-0.165) ]`

Explanation:

Using the following formulae :

`• r^2 = x^2 + y^2`

`• theta = tan^-1 (y/x)`

here x = 12 and y = - 2

hence `r^2 = 12^2 + (-2)^2 = 148 rArr r = sqrt148 = 2sqrt37`

and `theta = tan^-1(-2/12) ≈ -0.165" radians "`

`rArr (12-2i) = 2sqrt37 [cos(-0.165) + isin(-0.165) ]`