Complex Analysis Q&A - Singularities, Integration and More

[Niles]

Hi all.

I have some questions on complex analysis. They are very fundemental.

1) Singularities of a complex functions are the points, where the functions fails to be analytic. Will a singularity then always be a point, where the numerator of the functions is zero?

2) This question is on integration in the complex plane. If have a function f(z), then I have to specify the curve C (parametrized by z(t)), on which I wish to integrate f(z) along. If I just want to find the length of the curve, then which of the following integrals are correct?

<br /> \int dz \quad \quad \text{or}\quad \quad \int z(t) dz.<br />

I hope you can help. I would very much appreciate it.Niles.

 

[Niles]

Perhaps it would me more appropriate if this question was in the "Calculus and beyond"-homework forum. Would a moderator be kind enough to move the thread over there?

 

[ice109]

are you looking at this stuff because of the born scattering section in griffiths? lol

you are correct that a singularity of a function is a point where the function fails to be analytic but you are incorrect in saying that it's where the numerator is zero. for a rational function the singularities are where the denominator is zero. for other function it isn't necessarily so. for example y^2=x has a singularity at (0,0) because the slope is infinite there. |x| also has a singularity at (0,0).

for a line integral in complex space you need to specify a path z(t) where t is the parameter hence the path z is parameterized somehow by t.

in the integral it looks like this:

\int f(z)dz =\int f(z(t))\frac{dz}{dt}dt = \int f(z(t))z&#039;(t)dt

you can think of dz as a line element and hence the arc length is just

\int dz =\int z&#039;(t)dt

edit

actually that's wrong

since the line integral over complex space is the integral over a sort of vector field the definition i wrote down is correct. finding the arc length though corresponds to a line integral over a scalar field hence to find the arc length it should look like this:

\int |dz| =\int |z&#039;(t)|dt =\int \sqrt{z&#039;(t)*z&#039;(t)^*}dt = \int \sqrt{u&#039;(x(t),y(t))^2+v&#039;(x(t),y(t))^2}dt

where the star in the exponent of the second z'(t) in the middle exression is complex conjugation.

 

[Niles]

1) I actually meant the denominator (the lower part of the fraction), but I got the two terms mixed up.

2) So if I want to find the arc length of the unit circle, then it is given by:

<br /> \int_0^{2\pi} (\cos t + i\sin t)(-\sin t + i\cos t)dt ?<br />

Thanks.

 

[ice109]

1) I actually meant the denominator (the lower part of the fraction), but I got the two terms mixed up.

2) So if I want to find the arc length of the unit circle, then it is given by:

<br /> \int_0^{2\pi} (\cos t + i\sin t)(-\sin t + i\cos t)dt ?<br />

Thanks.

if the parameterization you're using is z(t)=\cos(t)+i\sin(t) then

z&#039;(t) = -\sin(t)+i\cos(t)
z&#039;(t)*= -\sint(t)-i\cos(t)
z&#039;(t)*z&#039;(t)= \sin^2(t) +i\cos(t)\sin(t)-i\sin(t)\cos(t)+\cos^2(t) =\sin^2(t) +cos^2(t)
|dz| =\sqrt{z&#039;(t)*z&#039;(t)}dt =\sqrt{\sin^2(t) +\cos^2(t)}dt = dt