Question

What is the trigonometric form of `(-10+3i)`?

Answer 1

`(-10+3i)=sqrt109(Cos(tan^-1(-3/10))+isin(tan^-1(-3/10)))`

Explanation:

If `(a+ib)` is a complex number, `u` is its magnitude and `alpha` is its angle then `(a+ib)` in trigonometric form is written as `u(cosalpha+isinalpha)`.
Magnitude of a complex number `(a+ib)` is given by`sqrt(a^2+b^2)` and its angle is given by `tan^-1(b/a)`

Let `r` be the magnitude of `(-10+3i)` and `theta` be its angle.
Magnitude of `(-10+3i)=sqrt((-10)^2+3^2)=sqrt(100+9)=sqrt109=r`
Angle of `(-10+3i)=Tan^-1(3/-10)=tan^-1(-3/10)=theta`

`implies (-10+3i)=r(Costheta+isintheta)`
`implies (-10+3i)=sqrt109(Cos(tan^-1(-3/10))+isin(tan^-1(-3/10)))`