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I would like to show that when a circular disk $|z| \leq \rho$ is translated one unit to the right, the point of maximum modulus in the resulting disk $|z+1| \leq \rho$ is $ z = 1 + \rho.$ Any hint or a proof for this?

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One unit to the left maybe? The point of maximum modulus in the disk you gave is $z=-1-\rho$. – copper.hat Nov 5 '12 at 20:23
Can you elaborate on this? – Zizo Nov 5 '12 at 20:25
If you move the first disk one unit to the right, the equation is $|z-1| \leq \rho$. – copper.hat Nov 5 '12 at 20:26
Another way to see this is to consider $|z+1|$ if $z=1+\rho$. Then $|z+1| = 2+\rho$, and clearly, if $\rho\geq 0$, we cannot have $2+\rho \leq \rho$. So, you need to change the disk or the point of maximum modulus. – copper.hat Nov 5 '12 at 21:19

Assuming you mean to find the maximum modulus $z$ in $C= \{ z | |z-1| \leq \rho\}$, you could note that if $z\in C$, $|z|-1 \leq |z-1| \leq \rho$. which gives $|z| \leq \rho +1$. Since $\hat{z} = 1+\rho \in C$, this gives $|z|\leq |\hat{z}| = |1+\rho| = 1+\rho$ for all $z \in C$.

To show that $\hat{z}$ is the only point of maximum modulus (assuming $\rho>0$), suppose $|z|=1+\rho$ and $|z-1| \leq \rho$. Letting $z=x+iy$, this gives $x^2+y^2 = 1 + 2 \rho + \rho^2$, and $x^2-2x+1 +y^2 \leq \rho^2$. Simplifying yields $\rho+1 \leq x$. Since $|z| = 1+\rho$, this gives $x^2 + y^2 \leq x^2$, from which it follows that $y=0$ and $x=\rho+1$. Hence $z=\hat{z}$.

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Hint: If $|z| \leq p$ then $|z+1| \leq |z| + 1 \leq p+1$, and if $z \neq p$ then at least one of these inequalities is strict.

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