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?
On
You may rearrange: \begin{align*} \vert a + b \mathrm{i} + 1 \vert = \vert b - a \mathrm{i} +1 \vert \\ \end{align*} Let's introduce $z = a+b \mathrm{i}$, the equation becomes \begin{align*} &\vert z + 1 \vert = \vert -\mathrm{i} z +1 \vert \\ &\vert z + 1 \vert = \vert - \mathrm{i} \vert \vert z + \mathrm{i} \vert \\ &\vert z + 1 \vert = \vert z + \mathrm{i} \vert \\ \end{align*} The equation in words: "find all complex numbers $z$ for which the distance in the complex plane to the point $-1$ is equal to the distance to the point $-\mathrm{i}$".
The solution is all points $z$ that lie on the line through the origin of the complex plane with slope $1$, i.e. all $z = a + b \mathrm{i}$ for which $a = b$.
On
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).
Hint:
Use the fact that for any complex number $z_1$ and $z_2$, $$\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$$ Then try and rearrange.