Like the title states, I'm trying to prove that $\left(\frac{z_1}{z_2}\right)^{\star} = \left(\frac{z_1^{\star}}{z_2^{\star}}\right)$ where z is a complex number and z* is its conjugate.
I keep finding myself stuck at $\left(z_2^{-1}\right)^{\star}$ in some form or another and can't find a valid reason why I can say that $\left(z_2^{-1}\right)^{\star} = \left(z_2^{\star}\right)^{-1}$.
My complex analysis books have what I'm trying to prove as a theorem but none of them actually go about proving it. I hope someone can point me in the right direction.
Thank you for your time.
To show that $(z^*)^{-1} = (z^{-1})^*$, you can use $z=a+ib$, where $(z^*)^{-1} = \frac{1}{a-ib} = \frac{a+ib}{a^{2} + b^{2}}$ and $(z^{-1})^* = (\frac{1}{a+ib})^* = (\frac{a-ib}{a^{2}+b^{2}})^*=\frac{a+ib}{a^{2}+b^{2}}$. Having this, can you prove the relation?