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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 $\bar{D}-interior(D)$?

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

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  • $\begingroup$ 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. $\endgroup$ – user1050421 Apr 26 '16 at 18:38

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