Question

If z = 2(`cospi/4+isinpi/4`), how do you write (-z) in mod-arg form?

Answer 1

Please see below.

Explanation:

If we have a complex number `z=a(costheta+isintheta)`,

then `-z=a(cos(pi+theta)+isin(pi+theta))`, where `-z` is additive inverse of `z`.

`barz=a(cos(-theta)+isin(-theta))`, where `barz` is conjugate of `z`.

Hence if `z=2(cos(pi/4)+isin(pi/4))`

`-z=2(cos((5p)/4)+isin((5pi)/4))`

and `barz=2(cos(-pi/4)+isin(-pi/4))`