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Principal Value and Principal Branch of an Integrand

Use parametric representation in exercise 10 for the oriented circle C0 there to show that....where a is any real number other than zero and where the principal branch of the integrand and where the principal value of R^G are taken.

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Ok the change of varibale is this:

z=z0+Rexp(i*theta) ---> dz=i*R*exp(i*theta)*dtheta and naturally:

-Pi < theta < Pi, so:

int((z-z0)^(a-1)dz)= ...

Solution Summary

An integral is solved using the principal value and principal branch of an integrand.