Integrating Complex Functions in the Complex Plane

dykuma
Messages
55
Reaction score
6

Homework Statement


Evaluate the following line integrals in the complex plane by direct integration.
upload_2016-11-13_21-56-9.png

Homework Equations


Z= x+i y = Cos(θ) +i Sin(θ) = e^i*θ

The Attempt at a Solution


I'm not sure how to evaluated this by hand. I tried using Z= x+i y = Cos(θ) +i Sin(θ), and evaluating the integral at dθ. However, I'm not sure how to change the bounds. It seems to me that point A starts at 2pi, and then point B is at 2pi + i *(infinity). what exactly does that mean?
 

Attachments

  • upload_2016-11-13_21-56-45.png
    upload_2016-11-13_21-56-45.png
    1.8 KB · Views: 432
  • upload_2016-11-13_21-56-48.png
    upload_2016-11-13_21-56-48.png
    1.8 KB · Views: 406
Physics news on Phys.org
dykuma said:

Homework Statement


Evaluate the following line integrals in the complex plane by direct integration.
View attachment 108889

Homework Equations


Z= x+i y = Cos(θ) +i Sin(θ) = e^i*θ

The Attempt at a Solution


I'm not sure how to evaluated this by hand. I tried using Z= x+i y = Cos(θ) +i Sin(θ), and evaluating the integral at dθ. However, I'm not sure how to change the bounds. It seems to me that point A starts at 2pi, and then point B is at 2pi + i *(infinity). what exactly does that mean?
You overcomplicate the problem. Do the integral with respect to z, as if it was a common real number. Then substitute the limits for z, using that the upper limit means z=x+iy=2pi + iy, y-->infinity.
 
  • Like
Likes dykuma
ehild said:
You overcomplicate the problem. Do the integral with respect to z, as if it was a common real number.Then substitute the limits for z, using that the upper limit means z=x+iy=2pi + iy, y-->infinity.
I see. That is what I wanted to do at first. However, I was taken back by the complex part of the upper bound. I see now that I really over complicated that problem.

Thank you!
 

Attachments

  • upload_2016-11-13_23-8-29.png
    upload_2016-11-13_23-8-29.png
    403 bytes · Views: 407
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top