I have been trying to evaluate the real and imaginary parts of $\frac{z}{(1-z)^2}$ and I was wondering if my solution was correct, as it's quite not nice.
I first wrote out the z's as (x+iy)'s, turning the denominator into $1+x^2 -y^2 -2x +i(2y(1+x))$. Then I multiplied the expression by the complex conjugate of the denominator and did a lot of simplification to turn the expression into $\frac{x^3-2x^2+x+xy^2+2xy+i(-y^3+y-yx^2-2yx-2x^2)}{x^4+y^4-4x^3+2y^2+6x^2-4x+12y^2+2y^2x^2+1}$ . Obviously here you can just split the fraction to obtain the real and imaginary parts. It just seems quite heavy, and I was wondering if my process wasn't the best for this type of problem. Thank you!
Best Answer
Here is a simpler method: Write $1-z=re^{i\theta}$. You get $\frac {1-re^{i\theta}} {r^{2}e^{2i\theta}}$. Split this into two terms and you will get the answer easily. [$\frac 1 {r^{2}}e^{-2i\theta}-\frac 1 re^{-i\theta}$ has real part $\frac 1 {r^{2}} \cos (2 \theta)-\frac 1 r \cos \theta$ and imaginary part $-\frac 1 {r^{2}} \sin (2 \theta)+\frac 1 r \sin \theta$. Use the identities $\cos (2\theta)=2\cos^{2}\theta -1$ and $\sin (2 \theta)=2 \sin \theta \cos \theta$].