Solve Complex Equation: Find Z, K in \frac{Z-a}{Z-b}=Ke^{±jθ}

[powers]

Hi, I need a little help

I need to find solution for this equations:

\frac{Z-a}{Z-b}=Ke^{±jθ}

The Z is unknown and it is the complex number. The a and b is known and they are also complex numbers. K is the real number.

I know that for -90^{°}<θ<90^{°} the graph in the complex plane is circle, for -45^{°}<θ<45^{°} the graph in the complex plane is in shape of "tomato" and for -135^{°}<θ<135^{°} is shape of "lens", but I don't know how to solve it.

Sorry if my post is in wrong area.

Thanks for help.

 

[jfgobin]

Leaving it to you the conditions of existence:

Z=\frac{a-b.K.\textrm{e}^{ \pm j \theta }}{1-K.\textrm{e}^{ \pm j \theta }}

 

[powers]

In that way I got only the one solution, where are the other?
For example, let's put b=0, K=1, theta=45°, with above formula we got only the one solution, but there is more than one solution...

 

[Char. Limit]

How do you only get one solution when there's clearly a \pm in his answer?