Real part of a complex number divided: $\Re\frac{z+1}{z-1}=0$

Question

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

I've tried so many methods, they all end up with two variables $a, b$.

I tried setting $z= a+ib$. This give me the equality $2\cdot\Re\frac{z+1}{z-1} = \frac{(a+1)+ib}{(a-1)+ib} + \frac{(a+1)-ib}{(a-1)-ib}$ which I'd then cross multiply.

I've tried so many methods and they all end up with something like $$\frac{1}{2}\times\frac{(a^2+b^2-2 i b-1)}{(a^2-2a+1+b^2)} = 0$$

And that makes no sense at all.

According to Wolfram, the answer to $\Re\frac{z+1}{z-1}=0$ is $-1$

Answer 1

Multiply and divide by the conjugate of the denominator to find $$\frac{z+1}{z-1} \cdot \frac{\bar z - 1}{\bar z -1} = \frac{(z+1)(\bar z -1)}{|z-1|^2} = \frac{z\bar z - z + \bar z - 1}{|z-1|^2}.$$

Now writing $z=x+iy$ we have:

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

Answer 2

We will start with the following transformations. We assume that $z\neq1$.

$$\begin{align} (z+1)-(z-1)=2\\ \frac{(z+1)(z-1)}{z-1}-(z-1)=2\\ \left(\frac{z+1}{z-1}-1\right)\cdot(z-1)=2 \end{align}$$

From here we get that

$$\frac{z+1}{z-1} = \frac{2}{z-1}+1.$$

Take the real part of both side.

$$\Re\left(\frac{z+1}{z-1}\right) = 2\,\Re\left(\frac{1}{z-1}\right)+1.$$

Now by definition of the real part

$$\Re\left(\frac{1}{z-1}\right)=\frac{1}{2}\left(\overline{\left(\frac{1}{z-1}\right)}+\left(\frac{1}{z-1}\right)\right).$$

Because the properties of the complex conjugate

$$\Re\left(\frac{1}{z-1}\right)=\frac{1}{2}\left(\left(\frac{1}{\overline z-1}\right)+\left(\frac{1}{z-1}\right)\right)=\frac{\overline z + z - 2}{2(z-1)(\overline z -1)}.$$

Now if we put it all together we get

$$\Re\left(\frac{z+1}{z-1}\right) = 2\,\Re\left(\frac{1}{z-1}\right)+1 = \frac{\overline z + z - 2}{(z-1)(\overline z -1)} + 1 = 0.$$

From here

$$\begin{align} \frac{\overline z + z - 2}{(z-1)(\overline z -1)} + 1 & = 0 \\ z - \overline z + 2 & = z\overline z + z - \overline z + 1 \\ z\overline z & = 1. \end{align}$$

So

$$\Re\left(\frac{z+1}{z-1}\right) = 0 \iff z\overline z = 1 \iff \left(\Re(z)\right)^2 +\left(\Im(z)\right)^2 = 1.$$