Question

`Arg(z) = -Arg(z^-1)` for any `zinC, z!=0` How to prove it?

Answer 1

Please see below.

Explanation:

Let `z=a+ib=rcostheta+irsintheta`

then `|z|=sqrt(a^2+b^2)` and `Arg(z)=tan^(-1)(b/a)`

Let `b/a=tanalpha` then `Argz=alpha`

Note that `z!=0=>` that both `a` and `b` cannot be together zero i.e. `a^2+b^2!=0`

As such `z^(-1)=1/(a+ib)=(a-ib)/((a+ib)(a-ib))`

= `(a-ib)/(a^2+b^2)=a/(a^2+b^2)-ib/(a^2+b^2)`

and `Arg(z^(-1))=tan^(-1)((-b/(a^2+b^2))/(a/(a^2+b^2)))=tan^(-1)(-b/a)`

= `tan^(-1)(-tanalpha)`

But as `tan(-alpha)=-tanalpha`

`Arg(z^(-1))=-alpha=-Arg(z)`