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What I have so far:

Assume $|z|= 1$ and $z\neq-1$, then $z=1$ or $z=i$ or $z=-i$.

If $z=1$, $Re(1-1)= 0$ as needed, but for trying to prove $z=i$ and $z=-i$ I get $Re(1-i) = 1$ and vice versa...

What am I doing wrong?

Nathanson
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Filip
  • 131
  • $|z|=1$ is the unit circle in the complex plane. $\pm 1, \pm i$ satisfies $|z|=1,$ but these are not the only complex numbers that satisfies this relation. There are uncountably many. – Krish Mar 25 '15 at 11:48

2 Answers2

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Put

$$z=x+iy\;,\;\;x,y,\in\Bbb R\stackrel{\text{given}}\implies x^2+y^2=1\;,\;\;x\neq-1$$

and then

$$\frac{1-z}{1+z}=\frac{\overbrace{\color{red}{1-|z|^2}}^{=0}+\overline z-z}{|1+z|^2}=-\frac {2y}{(x+1)^2+y^2}i\implies\text{Re}\,\left(\frac{1-z}{1+z}\right)=0$$

Timbuc
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$$\begin{align}2Re\left(\frac{1-z}{1+z}\right)&=\frac{1-z}{1+z}+\frac{1-\overline{z}}{1+\overline{z}}\\&=\frac{(1-z)(1+\overline{z})+(1-\overline{z})(1+z)}{(1+z)(1+\overline{z})}\\&=\frac{2-2z\overline{z}}{1+z+\overline{z}+z\overline{z}}\\&=0\end{align}$$

The last equality is because $z\overline{z}=|z|^2=1.$

The first thing you are doing wrong is to assume that $|z|=1$ is the same as $z=\pm1,\pm i$. There are other numbers that satisfy this condition, e.g. $\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$ and many others.

The second thing you are doing wrong is to compute only $Re(1-z)$. They are asking for $Re\left(\frac{1-z}{1+z}\right)$.

Nathanson
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