Should be a basic complex analysis question

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Homework Statement


Let f:C-> C be an entire bijection with a never zero derivative, then f(z)=az+b for a,b\in C

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The Attempt at a Solution


I'm not sure where to begin with this problem. The only ways I see to attack this are based on somehow showing that f' is bounded and then use Liouville's Theorem to conclude that f' is constant.

I don't know what role being a bijection plays in this, however I know it is necessary because z-> e^z is entire with a never zero derivative, but clearly does not have the form given.

Also, it clearly can't be a polynomial of degree greater than 1 because of f'\neq0 and the Fundamental Theorem of Algebra, but I don't know how to exclude a function of another form. I've been thinking about the Taylor series expansion of f, but I'm not sure why having an infinite number of terms be nonzero (otherwise it would be a polynomial) contradict either f' never zero or f being bijective.
 
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Try thinking about g(z)=f(1/z). g(z) has some kind of singularity at zero. Can you say what kind it can't be?
 
Interesting, it can't be an essential singularity because Picard's Theorem would then yield that f is not injective.

Thanks for the help!
 
Very welcome. A hint was all you needed. Great!
 
Ya, I just couldn't see how to connect the properties of f to reach the desired result until your hint.

Thanks again for the help.
 
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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