Plot The Area Of $Im(\frac{z-1}{z+1})=0$

Question

$Im(\frac{z-1}{z+1})=0$

$\frac{z-1}{z+1}=0$

$\frac{x-1+yi}{x+1-yi}=0$

$\frac{x-1+yi}{x+1-yi}\cdot \frac{x-1+yi}{x+1+yi}=0$

$\frac{x^2-1+y^2}{(x+1)^2+y^2}=0$

So

${x^2+y^2}=1$

which is circle of radius $1$?

Answer 1

Let $z$ be $x+iy$. Then,

$$\frac{z-1}{z+1}=\frac{x-1+yi}{x+1+yi}=\frac{x-1+yi}{x+1+yi}\cdot \frac{x+1-yi}{x+1-yi}=\frac{(x^2+y^2-1)+i(2y)}{(x+1)^2+y^2}$$ Since, $Im(\frac{z-1}{z+1})=0$ and $(x+1)^2+y^2\ne 0$ $$2y=0$$ Therefore, $y=0$. Hence, $f(x)=\frac{(x^2-1)}{(x+1)^2}=\frac{(x+1)(x-1)}{(x+1)^2}=\frac{(x-1)}{(x+1)}$, assuming $x \ne-1$.

Answer 2

Hint:   using that $\,w - \overline w = 2i \operatorname{Im}(w)\,$:

$$\require{cancel} \begin{align} 2i\operatorname{Im}\left(\frac{z-1}{z+1}\right)=\frac{z-1}{z+1} - \frac{\bar z-1}{\bar z+1} &= \frac{(z-1)(\bar z+1)-(z+1)(\bar z -1)}{|z+1|^2} \\ &= \frac{\bcancel{|z|^2}+z-\bar z-\cancel{1}-(\bcancel{|z|^2}-z+\bar z - \cancel{1})}{|z+1|^2} \\ &= \frac{2(z-\bar z)}{|z+1|^2} \\[5px] &= \frac{4i \operatorname{Im}(z)}{|z+1|^2} \end{align} $$