Complex Analysis - The Maximum Modulus Principle

smcro5
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Homework Statement



Find the maximum of \left|f\right| on the disc of radius 1 in the Complex Plane, for f(z)=3-\left|z\right|^{2}

Homework Equations



The maximum modulus principle?

The Attempt at a Solution



Since |z| is a real number, then surely the maximum must be 3 when z=0? However, I was reading that the maximum must occur on the boundary, which is |z|=1, for the disc which is described by |z|≤1. What am I doing wrong? Thanks in advance for any help!
 
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The Maximum Modulus Theorem applies to analytic functions. Is yours analytic?
 
Ah thanks for that, jackmell, much appreciated! It looks like 3-|z|^2 is not analytic, so the maximum modulus principle must therefore not apply. In situations like this though, we weren't taught how to deal with such things. So would my initial assumption be correct that the maximum of |f| is indeed 3 at z=0 ?
 
smcro5, u might be able to say that the function is analytic at x=y=0, so if u bound a domain there it should occur there, so yeah most likely the maximum should be 3. You can plot the function, It looks like a mexican hat ;)
 
Cheers Matty_t69, now it all makes sense! The plot of the function just makes the question a wee tad more exciting eh? ;)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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