What is the Indefinite Integral of \sqrt{z} along a Path in the Third Quadrant?

[mattmns]

Here is the exercise:

Use the indefinite integral to compute \int_{C} \sqrt{z}dz where C is a path from z = i to z = -1 and lying in the third quadrant. Note: \sqrt{z} = e^{(1/2)lnz} where the principal branch of lnz is defined on C \setminus [0,\infty].
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I am just a little unsure of why he gave us that note (although I do use it). Here is what I did for the exercise:

\int_{C} \sqrt{z}dz = \left[ \frac{2}{3}z^{3/2} \right]_{i}^{-1}

= \frac{2}{3} \left[ e^{(3/2)ln(-1)} - e^{(3/2)ln(i)} \right]

= \frac{2}{3} \left[ e^{(3/2)\pi i} - e^{(3/2)(\pi/2) i} \right]

= \frac{2}{3}(1 - 2i)

Everything look fine?

Thanks.

 

[StatusX]

I think you computed e^(3/2 pi/2 i) incorrectly, but other than that it looks ok. The reason he was explicit about the branch cut was because you'd get a different answer if the cut passed through the path.

 

[mattmns]

Woops, I forgot the whole \frac{\sqrt{2}}{2} part in there. Thanks!