What is the Contour Integral of Log(z) on a Specific Contour?

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SUMMARY

The contour integral of Log(z) over the specified contour defined by x^2 + 4y^2 = 4, where x >= 0 and y >= 0, requires parametrization of the contour as z(t) = 2cos(t) + isin(t) for 0 <= t <= pi/2. The integral is expressed as ∫Log(z(t))z'(t)dt, with Log(z(t)) calculated as ln|z(t)| + iArg(z(t)). The discussion highlights confusion regarding the use of the Fundamental Theorem of Calculus in evaluating the integral, particularly in determining the correct antiderivatives.

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Homework Statement


Find the contour integral of Log(z). The contour is defined as: x^2 + 4y^2 = 4, x>= 0, y>=0

Homework Equations




The Attempt at a Solution


parametrize the contour as z(t) = 2cos(t) + isin(t)
0 <= t <= pi/2
The contour integral = ∫Log(z(t))z'(t)dt
I am having trouble finding Log(z(t)).
Log(z(t)) = ln|z(t)| + iArg(z(t))

Would ln|z(t)| = ln|sqrt(1 + 3cos(t)^2)| and Arg(z(t)) = t?
 
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What's wrong with antiderivatives? I mean why not use the Fundamental Theorem of Calculus for this? You know the starting point of integration and the ending point so poke-a-poke right?
 

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