How to solve $\left|\frac{1 + a + bi}{1 + b - ai}\right| = 1$ I have a problem with solving following equation:
$$\left|\frac{1 + a + bi}{1 + b - ai}\right| = 1$$ (where $a$, $b$ are real numbers and $i$ is an imaginary unit) 
I tried to simplify its left side to something like $c + di$ but I don't know any method to achieve it in this case. Do you have any ideas how do it?

Write $z=a+bi$, so your equation be written as $$\left|\frac{1+z}{1-iz}\right| = 1$$ (verify this!). This is equivalent to $$|1+z| = |1-iz|$$ as long as we remember that $z\neq -i$ (to ensure a nonzero denominator in the original equation). The RHS can be written as $|-i(i+z)|$, and since $|-i|=1$, we can write the equation as
$$|z-(-1)| = |z-(-i)|$$
This equation  says that the solutions are the points in $\mathbb C$ that are equidistant from $-1$ and $-i$. These are precisely the points on the perpendicular bisector of the segment joining $-1$ and $-i$, which is the diagonal line through the origin, so the solutions are the points of the form $$\boxed{z = t + ti}$$ for real $t$. Note that $-i$ is not among them (if it were, we would have to exclude it because it would make the LHS of the original equation undefined).
