- #1
mattmns
- 1,128
- 6
Here is the exercise:
Use the indefinite integral to compute [tex] \int_{C} \sqrt{z}dz[/tex] where C is a path from z = i to z = -1 and lying in the third quadrant. Note: [tex] \sqrt{z} = e^{(1/2)lnz}[/tex] where the principal branch of lnz is defined on [tex]C \setminus [0,\infty][/tex].
-------
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:
[tex] \int_{C} \sqrt{z}dz = \left[ \frac{2}{3}z^{3/2} \right]_{i}^{-1} [/tex]
[tex]= \frac{2}{3} \left[ e^{(3/2)ln(-1)} - e^{(3/2)ln(i)} \right] [/tex]
[tex]= \frac{2}{3} \left[ e^{(3/2)\pi i} - e^{(3/2)(\pi/2) i} \right] [/tex]
[tex]= \frac{2}{3}(1 - 2i)[/tex]
Everything look fine?
Thanks.
Use the indefinite integral to compute [tex] \int_{C} \sqrt{z}dz[/tex] where C is a path from z = i to z = -1 and lying in the third quadrant. Note: [tex] \sqrt{z} = e^{(1/2)lnz}[/tex] where the principal branch of lnz is defined on [tex]C \setminus [0,\infty][/tex].
-------
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:
[tex] \int_{C} \sqrt{z}dz = \left[ \frac{2}{3}z^{3/2} \right]_{i}^{-1} [/tex]
[tex]= \frac{2}{3} \left[ e^{(3/2)ln(-1)} - e^{(3/2)ln(i)} \right] [/tex]
[tex]= \frac{2}{3} \left[ e^{(3/2)\pi i} - e^{(3/2)(\pi/2) i} \right] [/tex]
[tex]= \frac{2}{3}(1 - 2i)[/tex]
Everything look fine?
Thanks.
Last edited: