Question

If $Re\left(\frac{z-1}{z+1}\right)=0$, where $z=x+iy$ is a complex number, then which one of the following is correct?

• A. $z=1+i$
• B. $|z|=2$
• C. $z=1-i$
• D. $|z|=1$

Answer 1

The correct option is D $|z|=1$

Let $z=x+iy$

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

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

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

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

$=\frac{(x-1)(x+1)+(x-1)(-iy)+\left(iy\right)(x+1)+\left(iy\right)(-iy)}{(x+1{)}^2-(iy{)}^2}$

$=\frac{x^2-1+y^2+2yi}{x^2+2x+1+y^2}$

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

$Re\left(\frac{z-1}{z+1}\right)=0$

$\therefore Re\left(\frac{x^2-1+y^2+2yi}{x^2+2x+1+y^2}\right)=0$ ...(from 1)

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

$\therefore x^2-1+y^2=0$

$\therefore x^2+y^2=1$ ....(2)

$|z|=\sqrt{x^2+y^2}$

$=\sqrt{1}$ ...(from 2)

$=1$

So, option (D) is correct

Consider option (A)

$z=1+i$

$\therefore |z|=\sqrt{1^2+1^2}=\sqrt{2}$

So, option (A) is wrong.

Similarly it can be seen that option (C) is also wrong.

So, the only correct option is option (D)