Question

For complex numbers z and w, prove that ${{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w$, if and only if $z=w\ or\ z\overline{w}=1$.

Answer 1

Hint: We will be using the concept of complex numbers to solve the problem. We will be using the concept that modulus of a complex number ${{\left| z \right|}^{2}}=z\overline{z}$. Also if a complex number is real then $z=\overline{z}$.

Complete step-by-step answer:
Now, we have been given that we have to prove that ${{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w$ if and only if $z=w\ or\ z\overline{w}=1$.
Now, we will take the equation ${{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w$ and prove this is true if and only if $z=w\ or\ z\overline{w}=1$.
So, we have,
${{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w$
On rearranging terms, we have,
$\begin{align}
  & {{\left| z \right|}^{2}}w+w={{\left| w \right|}^{2}}z+z \\
 & w\left( 1+{{\left| z \right|}^{2}} \right)=z\left( 1+{{\left| w \right|}^{2}} \right) \\
 & \dfrac{w}{z}=\dfrac{1+{{\left| w \right|}^{2}}}{1+{{\left| z \right|}^{2}}} \\
\end{align}$
Now, since $\dfrac{1+{{\left| w \right|}^{2}}}{1+{{\left| z \right|}^{2}}}$ is purely real. Therefore, we have,
$\begin{align}
  & \left( \dfrac{w}{z} \right)=\left( \dfrac{\overline{w}}{z} \right)=\dfrac{\overline{w}}{z} \\
 & \dfrac{w}{z}=\dfrac{\overline{w}}{z} \\
 & w\overline{z}=z\overline{w}.......\left( 1 \right) \\
\end{align}$
Now, we again use the equation,
${{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w$
Now, we know that,
$\begin{align}
  & {{\left| z \right|}^{2}}=z\overline{z} \\
 & {{\left| w \right|}^{2}}=w\overline{w} \\
\end{align}$
So, we will use this in the equation and so we have,
$\begin{align}
  & z\overline{z}w-w\overline{w}z=z-w \\
 & \overline{z}zw-w\overline{w}z-z+w=0 \\
 & \overline{z}zw-z+w-w\overline{w}z=0 \\
\end{align}$
Now, we take z and w as common. So, we have,
$z\left( \overline{z}w-1 \right)-w\left( z\overline{w}-1 \right)=0$
Now, from (1) we have,
$\overline{z}w=z\overline{w}$
So, using this we have,
$\begin{align}
  & z\left( z\overline{w}-1 \right)-w\left( z\overline{w}-1 \right)=0 \\
 & \left( z-w \right)\left( z\overline{w}-1 \right)=0 \\
 & z=w\ or\ z\overline{w}=1 \\
\end{align}$
Hence, we have that,
${{\left| z \right|}^{2}}w-{{\left| w \right|}^{2}}z=z-w$ if and only if $z=w\ or\ z\overline{w}=1$.

Note: To solve these types of questions it is important to remember that if a complex number z is real that $z=\overline{z}$. Also few other identities to be remember like,
${{\left| z \right|}^{2}}=z\overline{z}$