Without (much) calculation find two loci

Question

The title pretty much says it. The problem I am trying to solve, is to find the loci of the expressions:

$$\text{a) $\operatorname{Re}\left(\frac{z-{(1+i)}}{z+(1+i)}\right)=0$}\quad \text{and}\quad \text{b) $\operatorname{Im}\left(\frac{z-{(1+i)}}{z+(1+i)}\right)=0$}$$

keeping the amount of calculation to a minimum.(For reference the problem can also be found in the "Art and Craft of problem solving" by Paul Zeitz p. 135 problem 4.2.9)

I really have no idea how to solve it, so any help would be greatly appreciated!

Answer 1

The first equally implies that the angle between $z-(1+i)$ and $z+(1+i)$ is $\pi/2$, implying that $z$ lies on the circumference of the circle whose diameter has endpoints $1+i$ and $-(1+i)$.

The second equality implies the angle between $z-(1+i)$ and $z+(1+i)$ is a multiple of $\pi$, implying that $z$ lies on the line joining $1+i$ and $-(1+i)$.

Answer 2

In the first case with $${\bf Re}\dfrac{z+a}{z-a}=\dfrac{|z|^2-|a|^2}{|z-a|^2}=0$$ shows $|z|=|a|$ where $z\neq a$. So $|z|=|-1-i|$ where $z\neq -1-i$ is a circle $|z|=\sqrt{2}$ except $z\neq -1-i$. Other case is similar.