Question

Let $w\ne \pm 1$ be a complex number. If $|w|=1$ and $z=\frac{w-1}{w+1}$, then $\text{Re}\left(z\right)$ is equal to

• A. $1$
• B. $\frac{1}{|w+1|}$
• C. $Re\left(w\right)$
• D. $0$
• E. $w+\overline{w}$

Answer 1

The correct option is D $0$

Given that:

$|w|=1$ and $z=\frac{w-1}{w+1}$

Let $w=x+iy$

Since, $|w|=1$ so, $\sqrt{x^2+y^2}=1\Rightarrow x^2+y^2=1$

Now,

$z=\frac{x+iy-1}{x+iy+1}$

$z=\frac{(x-1)+iy}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy}$

$z=\frac{x^2-1+ixy-ixy+2iy+y^2}{(x+1{)}^2+y^2}$

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

$Re\left(z\right)=\frac{x^2+y^2-1}{(x+1{)}^2+y^2}=0$ [$\because x^2+y^2=1$]