# $f(z)=\frac{1}{z^2-2z+2}$ - Maximum modulus principle

Let the function $f(z)=\frac{1}{z^2-2z+2}$. I have to find $\max_{z \in D(0,1)} |f(z)|$, but I already know that the maxixum would be on $\bar{D}-interior(D)$ by the maximum modulus principle. Is anyone could help me how to find the maximum on $\bar{D}-interior(D)$?

• Note that $z=e^{i\theta}=\cos\theta+i\sin\theta$ on the boundary and find the minimum of $\lvert 1/f(z)\rvert$. – user329501 Apr 26 '16 at 17:48
• You are absolutely right!!! – user1050421 Apr 26 '16 at 17:49

The question is equivalent to finding the minimum modulus of $(z-1)^2+1$ over $\|z\|\leq 1$.
That happens for $z=e^{i\theta}$: in such a case, $$\left\|(z-1)^2+1\right\|^2 = (\cos(2\theta)-2\cos\theta+1)^2+(\sin(2\theta)-2\sin\theta)^2$$ or: $$\left\|(z-1)^2+1\right\|^2 = 4(\cos^2\theta-\cos\theta)^2 + 4(\cos\theta-1)^2(1-\cos^2\theta)$$ or: $$\left\|(z-1)^2+1\right\|^2 = 16 \sin^4\frac{\theta}{2}.$$ The question is now trivial.
• So the minimum is zero, right? Because the derivative $(\sqrt{16 \sin^4\frac{\theta}{2}})'=2 \sin \theta =0 \implies \theta =0$. The answer of my teacher is $\sqrt{ \frac{1}{2}}$ for the minimum and $\sqrt{2}$ for the maximum, but I dont know where is my mistake. – user1050421 Apr 26 '16 at 18:38