A calculus of the absurd

14.1.5 Some example complex number problems

Example 14.1.2 Prove that
\(\seteqnumber{0}{14.}{24}\)
\begin{equation} \abs {z^{2022} e^{i \theta }} = \abs {\overline {z} e^{-i\theta }}^{2022} \end{equation}

where \(\overline {z}\) denotes the complex conjugate of \(z\).

We can prove this using the standard method for proving identities (see Section 6.2 for more details of the method) - pick one side and work to the other side using reversible steps.

\begin{align} |z^{2022} e^{i \theta }| &= |z^{2022}| |e^{i \theta }| \\ &= |(z^{*})^{2022}| |e^{i \theta }| \\ &= |(z^{*})^{2022}| |1| \\ &= |(z^{*})^{2022}| |1|^{2022} \\ &= |(z^{*})^{2022}| |e^{i \theta }|^{2022} \\ &= |(z^{*})^{2022}| |e^{-i \theta }|^{2022} \\ &= |(z^{*})|^{2022} |e^{-i \theta }|^{2022} \\ &= \Big (|(z^{*})| |e^{-i \theta }|\Big )^{2022} \\ &= |(z^{*})e^{-i \theta }|^{2022} \end{align}