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Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $

Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so

$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \cdot \frac{a+1-bi}{a+1-bi} = \frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a} = \frac{bi}{1+a} $$

and therefore $\ \frac{z-1}{z+1}$ is imaginary

now let me assume $\ \frac{z-1}{z+1} $ is imaginary number, how could I conclude that $\ |z| =1 $ I really can't think of any direction..

Thanks

bm1125
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    $$ \frac{(a^2+b^2)-1}{(a^2+b^2)+1 +2a}=0$$ – Nosrati Aug 29 '18 at 12:39
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    Multiply through by $\frac{z-1}{z-1}$ gives $\frac{(z-1)^2}{z^2-1}$ – Mandelbrot Aug 29 '18 at 12:39
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    Or also : https://math.stackexchange.com/questions/118868/for-complex-z-z-1-implies-textre-left-frac1-z1z-right-0 and https://math.stackexchange.com/questions/2548723/if-frac1-z1z-is-purely-imaginary-then-what-will-be-vert-z-vert – Arnaud D. Aug 29 '18 at 13:21

5 Answers5

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Note that\begin{align}\frac{z-1}{z+1}&=\frac{(z-1)\left(\overline z+1\right)}{(z+1)\left(\overline z+1\right)}\\&=\frac{|z|^2+z-\overline z-1}{|z+1|^2}\\&=\frac{|z|^2-1+2i\operatorname{Im}(z)}{|z+1|^2}\end{align}and that therefore $\frac{z-1}{z+1}$ is purely imaginary if and only $|z|^2-1=0$, which is the same thing as asserting that $|z|=1$.

José Carlos Santos
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Recall that $w\in \mathbb{C}$ is purely imaginary $\iff w=-\bar w$, then

$$\frac{z-1}{z+1}=-\overline{\left(\frac{z-1}{z+1}\right)} =-\frac{\bar z-1}{\bar z+1} \iff (z-1)(\bar z+1)=-(\bar z-1)(z+1)$$

$$z\bar z+z-\bar z-1=-z\bar z+z-\bar z+1$$

$$2z\bar z=2 \iff|z|^2=1$$

user
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In the last step of your proof, before you use the assumption that $(a^2+b^2)=1$, you've deduced

$$\frac{z-1}{z+1}=\frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a}$$

Now, by assumption for the other direction, you have

$$\frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a}=xi$$

as $\frac{z-1}{z+1}$ is assumed to be imaginary, i.e. $\frac{z-1}{z+1}=xi$ for some $x$. Thus,

$$\frac{(a^2+b^2)-1}{(a^2+b^2)+1 +2a}+\frac{2bi}{(a^2+b^2)+1 +2a}=xi$$

and therefore

$$\frac{(a^2+b^2)-1}{(a^2+b^2)+1 +2a}=0$$

i.e. $(a^2+b^2)-1=0$, and therefore$\sqrt{a^2+b^2}=1$.

blub
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Assume $z\ne\pm1$. Then ${z-1\over z+1}\in i{\mathbb R}$ means that, viewed from $z$, the two points $\pm1$ are seen under a right angle. By Thales' theorem this is the case iff $z$ is lying on a circle with diameter $[{-1},1]$.

Christian Blatter
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Another approach $${\bf Re}\dfrac{z-1}{z+1}={\bf Re}\dfrac{(z-1)(\bar{z}+1)}{|z+1|^2}=\dfrac{|z|^2-1}{|z+1|^2}=0 \iff |z|=1$$

Nosrati
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