Question

How do you write `6+0i` in trigonometric form?

Answer 1

`6(cos 0+i*sin 0)`

Explanation:

Trigonometric form of `a+bi` is `r(cos theta+sin theta)`

As such `r` is `sqrt(a^2+b^2)` and `theta=tan^(-1)(b/a)`

Here `a=6` and `b=0`, hence

`r=sqrt(6^2+0^2) = 6` and `theta=tan^(-1)0=0`

Hence `6+0i` in trigonometric form is `6(cos 0+i*sin 0)`.