# Complex Integral

1. Oct 17, 2006

### 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 $$C \setminus [0,\infty]$$.
-------

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.

Last edited: Oct 17, 2006
2. Oct 17, 2006

### 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.

3. Oct 17, 2006

### mattmns

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