1. The problem statement, all variables and given/known data
So, as far as I understand, there is not much difference between integrals in complex variables and those pesky line integrals from vector calculus. I'm not particullary good with line integrals, so it follows that I'm not quite understanding the complex variables ones.
Anyway, here is the problem
[itex]\oint (z^*)^2 dz [/itex] around the circunferences [itex] |z| = 1 [/itex] and [itex]|z-1|= 1[/itex] ([itex] z^*[/itex] is the complex conjugate.)
2. Relevant equations
[itex] z=x+iy [/itex]
[itex] f(z) = u(x,y) +iv(x,y) [/itex]
[itex] \int_c f(z)dz = \int_c (u dx -v dy) + i\int_c (v dx + u dy) [/itex] (1)
3. The attempt at a solution
First of all I use the Cauchy-Rienmann equations to see if the function is analytic inside the circuferences, so I could use Cauchy's theorem. Turned out, the function is analytic only at [itex]z=0[/itex]
[itex] f(z) = (x-iy)^2 = (x^2 - y^2) + i(-2xy) = u + iv [/itex]
[itex] \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \iff 2x = -2x \iff x = 0 [/itex]
[itex] \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \iff -2y = 2y \iff y = 0 [/itex]
so the function is analytic only on [itex] z=0 [/itex] and I have to integrate the hard way.
So I use (1) and now I have: [itex] \oint f(z)dz = \int_c (x^2 - y^2) dx +\int_c (2xy)dy + i\int_c (-2xy)dx +i\int_c (x^2 - y^2)dy [/itex]
so here is my problem: how do I integrate this?
what I tried is the following (and it might or migh not be correct);
using polar coordinates:
[itex] x =r\cos\theta[/itex]
[itex] y= r\sin\theta [/itex]
[itex]dx =-r\sin\theta d\theta [/itex]
[itex]dy =r\cos\theta d\theta [/itex]
the first integral [itex] \oint_c (x^2 - y^2)dx [/itex] becomes [itex] \int_0^{2\pi}( \cos^2 \theta - \sin^2\theta)(-sin\theta)d\theta = \int_0^{2\pi} (-\cos^2\theta sin\theta +sin^3\theta) d\theta [/itex] (remember that r = 1)
the idea is to do the same for the other integrals, integrate and then add.
I won't do it now because, like I said, I'm not sure this is the correct approach.
Can you you tell me if this would work or not?
So, as far as I understand, there is not much difference between integrals in complex variables and those pesky line integrals from vector calculus. I'm not particullary good with line integrals, so it follows that I'm not quite understanding the complex variables ones.
Anyway, here is the problem
[itex]\oint (z^*)^2 dz [/itex] around the circunferences [itex] |z| = 1 [/itex] and [itex]|z-1|= 1[/itex] ([itex] z^*[/itex] is the complex conjugate.)
2. Relevant equations
[itex] z=x+iy [/itex]
[itex] f(z) = u(x,y) +iv(x,y) [/itex]
[itex] \int_c f(z)dz = \int_c (u dx -v dy) + i\int_c (v dx + u dy) [/itex] (1)
3. The attempt at a solution
First of all I use the Cauchy-Rienmann equations to see if the function is analytic inside the circuferences, so I could use Cauchy's theorem. Turned out, the function is analytic only at [itex]z=0[/itex]
[itex] f(z) = (x-iy)^2 = (x^2 - y^2) + i(-2xy) = u + iv [/itex]
[itex] \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \iff 2x = -2x \iff x = 0 [/itex]
[itex] \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \iff -2y = 2y \iff y = 0 [/itex]
so the function is analytic only on [itex] z=0 [/itex] and I have to integrate the hard way.
So I use (1) and now I have: [itex] \oint f(z)dz = \int_c (x^2 - y^2) dx +\int_c (2xy)dy + i\int_c (-2xy)dx +i\int_c (x^2 - y^2)dy [/itex]
so here is my problem: how do I integrate this?
what I tried is the following (and it might or migh not be correct);
using polar coordinates:
[itex] x =r\cos\theta[/itex]
[itex] y= r\sin\theta [/itex]
[itex]dx =-r\sin\theta d\theta [/itex]
[itex]dy =r\cos\theta d\theta [/itex]
the first integral [itex] \oint_c (x^2 - y^2)dx [/itex] becomes [itex] \int_0^{2\pi}( \cos^2 \theta - \sin^2\theta)(-sin\theta)d\theta = \int_0^{2\pi} (-\cos^2\theta sin\theta +sin^3\theta) d\theta [/itex] (remember that r = 1)
the idea is to do the same for the other integrals, integrate and then add.
I won't do it now because, like I said, I'm not sure this is the correct approach.
Can you you tell me if this would work or not?
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