Question

prove that the conjugate of z1z2 is the same as the conjugate of z1 times the conjugate of z2

Answer 1

Let \(z_1=a+ib\) and \(z_2=c+id\), then
\[\begin{align*}
\bar{z}_1\bar{z}_2&=(a-ib)(c-id)\\
&=ac-i(ad+bc)+i^2bd\\
&=ac-bd-i(ad+bc)
\end{align*}\]
and
\[\begin{align*}
\overline{z_1z_2}&=\overline{(a+ib)(c+id)}\\
&=\overline{ac+i(bc+ad)+i^2bd}\\
&=\overline{ac-bd+i(bc+ad)}\\
&=ac-bd-i(bc+ad)\end{align*}\]
Thus \(\bar{z}_1\bar{z}_2=\overline{z_1z_2}\).

Answer 2

I'm using \(\bar{z}\) to denote the conjugate of \(z\), by the way.