I have an equation $|iz + 2 - i| = |\bar{z} - 4|$.
I tried to find a set of numbers and draw them on complex plane. I got something like $y = \frac{6x-11}{4}$. So, as I understand, all numbers will be on this line, right?
I also need to find all complex numbers which is a root of this equation and show them in trigonometric form. But I have no idea how to do it.
Lets clean these up to make them into a more familiar look
$iz + 2-i = i(z -(1+2i))\\ |i(z - (1+2i)| = |i||z -(1+2i)| = |z -(1+2i)|$
Since the real part of $\bar z$ equals the real part of $z$
$|\bar z -4 | = |z -4|$
We have
|z -(1+2i)| = |z -4|
$z$ is the set of points that lie on a line that perpendicularly bisects the segemet between the points $(4 + 0i)$ and $(1+2i)$
How to represent this line:
$\{x+yi | 6x - 4y = 11\}$
Or we can convert to polar:
$x = r\cos\theta\\y = r\sin \theta$
$\{r(\cos \theta + i\sin\theta) | r = \frac {11}{\cos\theta - r\sin \theta}\}$