Question about equation with complex variables

Question

I'm trying to solve this exercise:

Find all the complex numbers $z$ such that $\displaystyle w=\frac{z-1-i}{z+1+i}$ is a complex number that satisfies $|w|=1.$

The first thing that I did here was isolate $z$:

$$\displaystyle z=\frac{(w+1)+i\cdot(w+1)}{1-w}$$

If we define $w=a+bi$, the assumption is $a^2+b^2=1$, so we can isolate $a$ or $b$ and plug it into the expression of $z(w)$ to solve the system.

But I think that this is very ugly, so possibly there's another way to do this kind of exercises.

Thanks.

Answer 1

Hint. $|w|=1$ implies that $$|z-(1+i)|=|z+(1+i)|$$ that is $z$ has the same distance from $(1+i)$ and $-(1+i)$. What is the locus of such complex number $z$?

P.S. You may also let $z=x+iy$ and then solve $$(x-1)^2+(y-1)^2=|z-(1+i)|^2=|z+(1+i)|^2=(x+1)^2+(y+1)^2.$$

Answer 2

Hint...apply the condition $|w|=1$ directly so you have $$|z-i-1|=|z+1+i|$$

If you put $z=x+iy$ you will get a straight line

Answer 3

Well, the thing you need to solve is:

$$\left|\frac{\text{z}-1-i}{\text{z}+1+i}\right|=1\space\Longleftrightarrow\space\Re\left(\text{z}\right)+\Im\left(\text{z}\right)=0\tag1$$

Answer 4

WLOG $w=\cos t+i\sin t=e^{it}$

$$\dfrac{z-(i+1)}{z+(1+i)}=e^{it}$$

$$\dfrac z{1+i}=\dfrac{e^{it}+1}{1-e^{it}}=\dfrac{\cos t}{-i\sin t}$$

$$z=(1+i)i\cot t=-\cot t+i\cot t$$

If $z=x+iy,$ where $x,y$ are real

$x=-\cot t,y=\cot t\implies $ either $x=y=0$ or $\dfrac yx=?$