Complex number problem becomes huge when using formulas - is there any workaround?

Question

Here is an equation I need to solve for $z$ where $z$ is a complex number.(I need to show which complex numbers are solution for this problem): $$\left|\frac{1+z}{1-i\bar z}\right| = 1$$ Here are formulas I am using: $|a+bi| = \sqrt{a^2+b^2}$; $\bar z = a-bi$. So at first I need to simplify $\frac{1+z}{1-i\bar z}$ to $a+bi$ form. After a few operations here is what I came up with: $$\frac{1+z}{1-i\bar z} = \frac{1+a-b-2ab+(b-b^2+a+a^2)i}{1+b^2+2b+a^2}$$ where $z = a+bi$. It seems overcomplicated already and I still have to find modulus of this number which will complicate it even more. The question is - in such a case should I try to count this that way anyway or is there any simpler solution which I haven't noticed?

Answer 1

Given two complex numbers $z_1\ne z_2$, the equation $|z-z_1|=|z-z_2|$ says that the distance between $z$ and $z_1$ is equal to the distance between $z$ and $z_2$; the solution is the perpendicular bisector of the line segment joining $z_1$ and $z_2$. With a little easy work, your equation can be put in that form.

First, clearing fractions gets the equation in the form $$|1+z|=|1-i\bar z|.$$ The left side is $$|1+z|=|z-(-1)|,$$ the distance from $z$ to $-1$. The right side needs a bit more work. Using the identities $|zw|=|z||w|$ and $|\bar z|=|z|$, we get $$|1-i\bar z|=|\overline{1-i\bar z}|=|1+iz|=|i(-i+z)|=|i||-i+z|=1\cdot|z-i|=|z-i|.$$ So your equation simplifies to $$|z-(-1)|=|z-i|,$$ and the solution set is the set of all points equidistant from $-1$ and $i$, which is the line $y=-x$.

Answer 2

If $z = x+iy$ then your equation is (see Daniel's hint) $$ (1+x)^2+y^2 = (1+y)^2 + x^2 $$

So the solutions is $$ z = r (1+i) $$ where $r$ can be any real number.