Loser66
  • Loser66
Complex analysis problem Find \(\int_\gamma z^{-1/2} dz\) where \(\gamma \) is upper half of unit circle from 1 to -1 Please, help
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i forget don't you so something like put \(z(t)=e^{it}\) so \(z'(t)=ie^{it}\) ?
Loser66
  • Loser66
I have to use branch cut, my Prof said I can see z^(-1/2) as principal branch cut of z^(-1/2). I don't get what he meant
Loser66
  • Loser66
\(z^(-1/2) = e^{(-1/2) logz}\)

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anonymous
  • anonymous
is upper half of unit circle from 1 to -1 means this right?|dw:1447556104302:dw|
Loser66
  • Loser66
Yes
Loser66
  • Loser66
But when using branch cut, the log is undefined on negative of real part, we don't have -1 on the curve, how to argue?
Loser66
  • Loser66
|dw:1447556231668:dw|
Loser66
  • Loser66
|dw:1447556286628:dw|
anonymous
  • anonymous
guess you are using the top one right?
anonymous
  • anonymous
ooh i see the problem
Loser66
  • Loser66
why? it has 2 values, how can I ignore the second one? is it not that we must add them together?
anonymous
  • anonymous
i guess we can check
Loser66
  • Loser66
Show me, please
anonymous
  • anonymous
but that was wrong sorry
anonymous
  • anonymous
this is right i think http://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt%28exp%28i+t%29%29+i+exp%28i+t%29+df+from+t%3D0+to+pi
anonymous
  • anonymous
lets work it out if we can
Loser66
  • Loser66
I don't get why you put sqrt at the denominator. Is it not that \(\dfrac{1}{\sqrt z}= \dfrac{1}{e^{it}}\)
anonymous
  • anonymous
nope \[z^{-\frac{1}{2}}=\frac{1}{\sqrt{z}}\]
Loser66
  • Loser66
yes, replace z by e
anonymous
  • anonymous
yeah get \[\frac{1}{\sqrt{e^{it}}}\]
Loser66
  • Loser66
oh, got it. :)
anonymous
  • anonymous
but that is probably not how you are supposed to do it (maybe)
anonymous
  • anonymous
although i am not sure why not, so maybe i should shut up seems like it ought to work that way
Loser66
  • Loser66
Anyway, if we know the answer, it is way better than nothing :) Thanks a lot.
anonymous
  • anonymous
i think it is not so bad you end up with \[i\int_0^{\pi}e^{\frac{it}{2}}dt\]
Loser66
  • Loser66
Ok, let me try.
anonymous
  • anonymous
ok i am really doing this a basic way subtracting the exponents etc find the anti derivative , plug in \(\pi\) and \(0\) i remember doing it this way without writing as the log etc

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