Question

Prove that for any complex number `z`, `mod(1/z)=1/(modz)` and `Arg(1/z)=-Argz`?

Answer 1

Please see below.

Explanation:

Let `z=a+ib` and in polar form we can write it as

`z=re^(itheta)`, where `r=sqrt(a^2+b^2)` is modulus and `theta=tan^(-1)(b/a)` is Argand.

Hence, `1/z=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)`

Hence, while `mod(1/z)=sqrt((a/(a^2+b^2))^2+(b/(a^2+b^2))^2)`

= `1/sqrt(a^2+b^2)=1/(modz)`

This will be true is `a!=0!=b`

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

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

This is true if `a!=0!=b`