Find all complex numbers z, for which is $(\frac{(z - i - 1)}{(iz + 1)})^2$ real number?

Question

When I put in $z = x + iy,$ in the end I got: $${x^2 - 2x - y^2 + 2y + i (2 - 2y - 2x + 2xy) \over -x^2 + y^2 - 2y + 1 + i (2x - 2xy)}$$ Imaginary part is equal to $0$ and when I did that in the end I got: $x - xy + y = 1$ and $x (1-y) = 0.$ From the last equation I got one complex number but I do not know how to get others.

Answer is

$$z = x + iy, \quad (x - 1/2)^2 + (y - 1)^2 = 1/4,\quad x \ne 0$$

and

$$z = x + i,\quad x \ne 0$$

I want to get to this answer.

Thank you for all your answers.

Answer 1

• You want $\frac{z-i-1}{iz+1}$ to be real or imaginary (i.e., $t$ or $it$ with $t\in\Bbb R$).
• Note that $z\mapsto \frac{dz-b}{-cd+a}$ is an inverse map to $z\mapsto \frac{az+b}{cz+d}$

Answer 2

Since $w=\frac{z-i-1}{iz+1}\iff (1-iw)z=w+i+1$ and we want $w$ to be real or imaginary, we get a contradiction for $w=-i$ but otherwise obtain $z=\frac{w+i+1}{1-iw}$. There are two cases. In the first, $w$ is real so $z=\frac{(1+w+i)(1+iw)}{1+w^2}$ is the usual Cartesian form. In the second, write $w=ix$ with real $x\ne-1$, giving $z=\frac{1+(1+x)i}{1+x}$.

Answer 3

It suffices to let $\frac{z-i-1}{iz+1}=\frac{(z-i-1)(-i\bar z+1)}{|iz+1|^2}$ to be real or purly imaginary, i.e. $(z-i-1)(-i\bar z+1)$ to be real or purly imaginary.

Write $z=x+iy$ to obtain $(z-i-1)(-i\bar z+1)=y-1-[(y-1)^2+x(x-1)]i.$ So $y=1$ or $(y-1)^2+x(x-1)=0$, removing the case $x=0$ gives your answer.