- #1
strangequark
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Homework Statement
Compute the following integrals using the principle value of [tex]z^{i}[/tex]
a.
[tex]\int z^{i} dz [/tex] where [tex]\gamma_{1}(t)=e^{it}[/tex] and [tex]\frac{-\pi}{2}\leq t \leq \frac{\pi}{2} [/tex]
b.
[tex]\int z^{i} dz [/tex] where [tex]\gamma_{1}(t)=e^{it}[/tex] and [tex]\frac{\pi}{2}\leq t \leq \frac{3\pi}{2}[/tex]
Homework Equations
The Attempt at a Solution
There is a "hint" with the problem that says one of the integrals is easier than the other.
I don't see why, for part a, I can't use a branch cut along the negative real axis, so that [tex]z^{i}[/tex] will be analytic along the path.
And for part b, i don't see why I can't simply use a different branch cut, say one along the positive real axis, so that [tex]z^{i}[/tex] will then be analytic along that path.
Then for each, I can just take the antiderivative:
[tex]F(z)=\frac{z^{i+1}}{i+1}[/tex]
and plug in the end points...
If I do, I get:
a. [tex]\int z^{i} dz = \frac{i^{i+1}}{i+1}+\frac{(-i)^{i+1}}{i+1} = (\frac{e^{\frac{\pi}{2}}}{2}+\frac{e^{\frac{\pi}{2}}}{2}i)-(-\frac{e^{\frac{\pi}{2}}}{2}-\frac{e^{\frac{\pi}{2}}}{2}i ) = cosh(\frac{\pi}{2})+cosh(\frac{\pi}{2})i[/tex]
and I will get the same thing for b.
Is there something I am missing? Do I ned to parameterize the integral or is what I am doing correct?
Thanks in advance!
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