Conformal Mapping: Finding \phi(z) = z^{0.5}

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Homework Statement


Hi

Is there a rigorous way to find conformal mappins? Say I would like to find how \phi(z)=z^{0.5} maps the domain r\exp(i\phi) (with r>0 and 0\leq \phi \leq \pi), how would I do this?

Thanks in advance.
 
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Niles said:

Homework Statement


Hi

Is there a rigorous way to find conformal mappins? Say I would like to find how \phi(z)=z^{0.5} maps the domain r\exp(i\phi) (with r>0 and 0\leq \phi \leq \pi), how would I do this?

Please don't use the same symbol for two different objects in the same context! Either \phi is a complex function or it's the argument of a complex number. Choose one and stick with it, and find a different symbol for the other.

To answer your question: Start with \phi(re^{i\theta}) = r^{1/2}e^{i\theta/2}. What values can r^{1/2} take if r > 0? What values can \theta/2 take if 0 \leq \theta \leq \pi? What region of the complex plane does that give you?
 
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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