Find the polar equation of the loci of $\Im(z-1+\frac{4}{z})=0$ Question: Find the polar equation of the loci $$\Im(z-1+\frac{4}{z})=0$$
Where $\Im()$ means the imaginary part of the resulting complex number.
My Workings: Let $z = x+iy$
So that $$\Im(x+iy-1+\frac{4}{x+iy})=0$$
Therefore $$\Im((x-1)+iy+\frac{4(x-iy)}{x^2+y^2})=0$$
Therefore $$\Im(x-1+\frac{4x}{x^2+y^2} + i(y-\frac{4y}{x^2+y^2}))=0$$
This implies $$ y-\frac{4y}{x^2+y^2} = 0 $$
Now, I am stuck and don't know what to do.
 A: 
This implies $$ y-\frac{4y}{x^2+y^2} = 0 \tag1$$


Now, I am stuck and don't know what to do.

I haven't scrutinized your work that closely, so, there might be an underlying analytical error.  Assuming not, you have accomplished $90$% of the work needed.  Therefore, I upvoted your question.
Assuming that the above result is accurate, the problem is completed as follows:
$\underline{\text{Case 1:} ~y = 0.}$
Then, in order for the expression in (1) above to be meaningful, you can not have $x = 0$.  Futher, it is obvious that when $y = 0$ and $x$ is any non-zero value, then the equation in (1) above holds.
$\underline{\text{Case 2:} ~y \neq 0.}$
Divide by $y$ to produce
$\displaystyle 1-\frac{4}{x^2+y^2} = 0 \implies x^2 + y^2 = 4.$
Since $\displaystyle |z| = \sqrt{x^2 + y^2}$, the constraint immediately above
is equivalent to the constraint: $|z| = 2.$
Edit
Minor logic flaw in my Case 2 analysis.
Technically, the Case 2 result should be :
$y \neq 0$ and $|z| = 2$.
However, since the specific situation of $y = 0$ and $|z| = 2$ 
falls within the umbrella of Case 1, the logic flaw is harmless.
A: Either $z$ is non-zero real (in which case $z-1+4/z$ is also real and the polar equation is $\theta=0$) or
$$y=\frac{4y}{x^2+y^2}\implies1=\frac4{r^2}\implies r=2$$
A: Once you have the Cartesian equation, simply plug in the conversions $x=r\cos\theta$ and $y=r\sin\theta$ and see whether/how it may be simplified. In this case you shoukd ge able to see a factor of $r\sin\theta$ (or a factor of $y$ in the Cartesian form), and moreover the sum $x^2+y^2$ may be seen as $r^2$. You should get $(r+2)(r-2)\sin\theta=0$ if you want the whole thing represented by one equation.
For your further thinking: why do we have factors of both $r+2$ and $r-2$ when a graph of the relation shows just one circle?
Secondary point: I edited your question by introducing $\Im$ for the imaginary part. This and $\Re$ for the real part are widely used here.
