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:

$\frac{Z-a}{Z-b}$=K$e^{±jθ}$

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?

 

[CellCoree]

To solve this complex equation, we can begin by multiplying both sides by (Z-b) to eliminate the fraction. This gives us:

Z-a=K(Z-b)e^{±jθ}

Next, we can distribute the K and the e^{±jθ} terms to get:

Z-a=KZe^{±jθ}-Kbe^{±jθ}

Now, we can rearrange the equation to isolate Z on one side:

Z-KZe^{±jθ}=a-Kbe^{±jθ}

Next, we can factor out Z from the left side:

Z(1-Ke^{±jθ})=a-Kbe^{±jθ}

Finally, we can solve for Z by dividing both sides by (1-Ke^{±jθ}):

Z=\frac{a-Kbe^{±jθ}}{1-Ke^{±jθ}}

This equation gives us the solution for Z in terms of a, b, K, and θ. To find the specific values for Z and K, we will need to plug in the given values for a, b, and θ. This will give us a complex number for Z and a real number for K.

I hope this helps and feel free to ask for any clarification. Remember, solving complex equations can be tricky, so always double check your work and make sure to follow the order of operations. Good luck!