Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

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 $z=\hat{z}$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.