Proving Complex Number Equality

[squenshl]

Homework Statement

$z$ is a complex number such that $z = \frac{a+bi}{a-bi}$, where $a$ and $b$ are real numbers. Prove that $\frac{z^2+1}{2z} = \frac{a^2-b^2}{a^2+b^2}$.

Homework Equations

The Attempt at a Solution

I calculated
\begin{equation*}

\begin{split}

z = \frac{a+bi}{a-bi} &= \frac{a+bi}{a-bi}\times \frac{a+bi}{a+bi} \\

&= \frac{a^2+2abi-b^2}{a^2+b^2} \\

&= \frac{a^2-b^2}{a^2+b^2}+\frac{2ab}{a^2+b^2}i.

\end{split}

\end{equation*}
But sticking that ugly thing into $\frac{z^2+1}{2z}$ gives me something nasty. I'm sure there is a much simpler way!

 

[PeroK]

Homework Statement

$z$ is a complex number such that $z = \frac{a+bi}{a-bi}$, where $a$ and $b$ are real numbers. Prove that $\frac{z^2+1}{2z} = \frac{a^2-b^2}{a^2+b^2}$.

Homework Equations

The Attempt at a Solution

I calculated
\begin{equation*}

\begin{split}

z = \frac{a+bi}{a-bi} &= \frac{a+bi}{a-bi}\times \frac{a+bi}{a+bi} \\

&= \frac{a^2+2abi-b^2}{a^2+b^2} \\

&= \frac{a^2-b^2}{a^2+b^2}+\frac{2ab}{a^2+b^2}i.

\end{split}

\end{equation*}
But sticking that ugly thing into $\frac{z^2+1}{2z}$ gives me something nasty. I'm sure there is a much simpler way!

What could you do with $\frac{z^2+1}{2z}$?

 

[kuruman]

It might be prettier if you defined $u=a+bi$ in which case $z=u/u^*$. Put that ratio in $\frac{z^2+1}{2z}$ and see what you get.

 

[squenshl]

Great thank you very much!

 

[haruspex]

Another way might be by noting that $\frac {z^2+1}{2z}=\frac 1{\frac 1{z+i}+\frac 1{z-i}}$

 

[PeroK]

I think, as aluded to in post #2, the key is:

$\frac{z^2+1}{z}=z+\frac{1}{z}$