<s>[INST] Complex analysis computation of $Im(f(z)-f(z_0))$ I have a computation that I wanted to carry out and know the answer. My issue is that when I did the computation I did not arrive at the answer that I know to be true. I would really appreciate a fresh set of eyes to comment on my work.
For $f(z)= kz- \sqrt{z}$ we want to compute Im$[f(z)-f\left(\frac{1}{4k^2}\right)]=0$, where $k$ is a real number. What we should obtain is a parabola $x= \frac{1}{4k^2}-k^2y^2$.
My computations
Consider that Im$[f(z)-f\left(\frac{1}{4k^2}\right)]=0$ is the same as Im$[f(z)]=$Im$[f\left(\frac{1}{4k^2}\right)]$.
Then I compute
$f(\frac{1}{4k^2})=k \frac{1}{4k^2} - \sqrt{\frac{1}{4k^2}} =\frac{1}{4k}- \frac{1}{2k}= -\frac{1}{4k}$
Where we know that this is real thus we know the Im$[f\left(\frac{1}{4k^2}\right)]=0$
To compute Im$[f(z)]=0$ we let $z= x+iy$ then we observe
Im $[ k(x+iy) -\sqrt{x+iy}]=0$
Which is equivalent to Im$[k(x+iy)]=$Im$[\sqrt{x+iy}]$. Then to deal with the sqrt I use the exponential definition of complex numbers and see that $r= \sqrt{x^2+y^2}$ and $\theta$ be the corresponding angle. Thus Im$[k(x+iy)]=$Im$[\sqrt{x+iy}]$ is the same as $ky=$Im$[r^{1/2} e^{\theta/2}]$.
Then using Eulers identity to make $e^{\theta/2} =\cos \theta/2+i \sin \theta/2$.
Then we get that 
$$ky= r^{1/2} \cos\theta/2.$$
Squaring both sides we obtain 
$$k^2y^2= r \cos^2 \theta/2.$$
Using trig we get 
$$k^2y^2=\frac{1}{2} r(1+ \cos \theta).$$
 Now using that $\cos \theta = \frac{x}{r}$ we get 
$$k^2y^2 = \frac{1}{2}(r+ x).$$
We can turn $r$ into $x$ and $y$,
$$k^2y^2= \frac{1}{2}(\sqrt{y^2+x^2} +x).$$
However that is not the answer that is given to be true, i.e. $x= \frac{1}{4k^2}-k^2y^2$. I would really appreciate comments and solutions.
 [/INST] A mistake I see is that the imaginary part of $r^{1/2}e^{\theta/2}$ is $r^{1/2}\sin\theta/2$ (and not cosine). The change is minor, though: your last equality changes to 
$$
k^2y^2= \frac{1}{2}(\sqrt{y^2+x^2} -x).
$$
So 
$$
\sqrt{x^2+y^2}=x+2k^2y^2.
$$
Squaring,
$$
x^2+y^2=x^2+4xk^2y^2+4k^4y^4,
$$
so, after subtracting $x^2$ and dividing by $y^2$,
$$
1=4xk^2+4k^4y^2.
$$
Solving for $x$,
$$
x=\frac1{4k^2}-k^2y^2.
$$
</s>