Zeroes of derivative of analytic function Let $f(z) = z \prod\frac{z-a_i}{z-\bar a_i}$ where $\#\{a_i\} < \infty$ $Re\: a_i > 0\: Im\:a_i > 0$. I need to prove that $f'$ has zero $a:Re\: a > 0\: Im\:a > 0$. My attempts were to find argument of derivative among large enough square path starting from $0$ and going around old zeroes and use somehow argument principle. I have a thought that $f'$ has double amount of poles under the real line at same place  and we can somehow use it but i have zero experience in using these facts so I will be thankful for hints or directions.
 A: Sketch of the proof: 
Let $N$ the number of $a_k=x_k+iy_k, x_k, y_k >0$ and assume they are distinct wlog as otherwise, the result in the OP is immediate (though the method still works since only the behavior of the critical points corresponding to the poles of $f$ so the ones in the lower half-plane are affected by some $a_k$ coinciding) 
By the Theorem of Bocher-Walsh which shows that if $f=P/Q, \deg P=p \ge 1, \deg Q=q, p \ne q$ and $P$ has all zeroes in a half-plane $\Lambda$, $Q$ has all zeroes in a disjoint half plane $\Lambda_1$ (by continuity one of the half planes can be taken closed as here), then $f$ has $p-1$ critical points in the half plane $\Lambda$ and all the rest of the critical points ($q$ if the zeroes of $Q$ are distinct) in $\Lambda_1$ it follows that $f$ has $N$ critical points in the closed upper half-plane and $N$ in the open lower half-plane (though as we will easily see no critical point can be real anyway)
However here the result follows easily from the argument principle since writing $A=\prod (z-a_k), B=\prod (z-\bar a_k)$ and using that the $a_k$ are distinct we get that $f'(z)=0$ iff $T=AB+z(A'B-AB')=0, \deg T=2N$ and if $AB=P_0, z(A'B-AB')=iP_1, P_0,P_1$ have real coefficients and $P_0(x)>0, x \in \mathbb R$ so integrating $T'/T$ on the contour formed by the segment from $-R$ to $R$ and the half circle of radius $R$ in the upper half plane, we immediately get the number of zeroes in the upper half plane of $T$ is $N$ as the half circle integral obviously is about $N$ for large $R$, while the real line integral goes to zero since $\Re T >0$. 
But now dividing by $AB$ we have that the critical points $w$ of $f$ satisfy:
$1+\sum w(\frac{1}{w-a_k}-\frac{1}{w-\bar a_k})=0$ or equivalently
$1+\sum (\frac{a_k}{w-a_k}-\frac{\bar a_k}{w-\bar a_k})=0$
But if $w=-c+id, c,d \ge 0$ we easily see that $|w-a_k| <|w-\bar a_k|$ (that is true for any $w$ in the upper half plane actually regardless of the qudrant) and then $\Im \frac{a_k}{w-a_k}=\frac{1}{|w-a_k|^2}\Im (a_k \bar w)=\frac{-x_kd-y_kc}{|w-a_k|^2}$ and similarly for $\Im \frac{\bar a_k}{w-\bar a_k}=\frac{-x_kd+y_kc}{|w-\bar a_k|^2}$ which gives that $\sum \frac{x_kd+y_kc}{|w-a_k|^2}=\sum \frac{x_kd-y_kc}{|w-\bar a_k|^2}$ but LHS is bigger term by term than the absolute value of RHS and we cannot have equality since that implies $c=d=0$ and that is not possible since $0$ is not a critical point (also the real parts then do not match).
So all the critical points in the upper half-plane are in the first quadrant (strictly by the above as $c=0$ or $d=0$ implies $c=d=0$ as we saw) and since there are $N \ge 1$ of them we are done!
