Complex Line Integral (should be easy)

In summary: Thanks! In summary, the homework statement is trying to find the antiderivative of a function with an argument within the range -\pi/2 to \pi. It seems to be getting complicated, and there are many unknowns. Suggestions?
  • #1
Jamin2112
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12

Homework Statement



Using a partial fraction decomposition, show that if z lies in the right half plane and C is the line segment from 0 to z, then

C dz/(z2+1) = i/2 Log(z+i) - i/2 Log(z-i) + π/2​

Homework Equations



Log(z) = ln(z) + i Arg(z) (maybe relevant?)

The Attempt at a Solution



C dz/(z2+1) = -1/2 ∫C dz/(z+1) + 1/2 ∫C dz/(z-1) = -1/2 Log(z+1) + 1/2 Log(z+1) ... This isn't looking promising. Suggestions?
 
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  • #2
You're not doing the partial fractions correctly. You got i's in there you know. Factor z^2+1 and get:

[tex]\frac{1}{z^2+1}=\frac{1}{(z+i)(z-i)}[/tex]

now do the partial fractions and then consider the analyticity of the antiderivative along the integration path. Did I just say that? Jesus, what does that even mean? Where are the branch cuts in all of this? branch-points? Do we even need them? Why don't we just try an easier one first like just:

[tex]\int_0^1 \frac{dz}{z-i}[/tex]

that 's easy right? It's just:

[tex]\int_0^1 \frac{dz}{z-i}=\text{mylog}(z-i)\biggr|_0^1[/tex]

with mylog just being an analytic part of the multivalued function log(z-i). Say we define it as:

[tex]\text{mylog}(z-i)=\ln|z-i|+i\theta,\quad -\pi<\theta\leq \pi[/tex]

where theta is the argument (within that range above, of the quantity z-i. So z starts at 0 so then the argument of (0-i) is just -\pi/2. That's within range. Now say z=1/4 +1/4 i so arg of 1/4-3/4 i is what? I don't know exactly. You figure it out but it's in the fourth quadrant so still in range. Same way with the entire length of the integration path: the argument is always within that range above and never crosses the branch-cut along the negative real axis so mylog is analytic along the entire path of integration so therefore I can evaluate the integral by evaluating the antiderivative at it's end points and write:

[tex]\int_0^1 \frac{dz}{z-i}=\text{mylog}(z-i)\biggr|_0^1=\text{mylog}(1-i)-\text{mylog}(-i)[/tex]

You not gettin' confussed with that mylog thing right? It's just log ok, or rather the piece of it I define above. You can do the other one. Also, see this thread about using parts of log. Make sure you understand what's going on.

https://www.physicsforums.com/showthread.php?t=552090
 
Last edited:
  • #3
I see what's going on now!
 

What is a complex line integral?

A complex line integral is a mathematical concept used in vector calculus to calculate the work done by a vector field along a curve in the complex plane.

How is a complex line integral calculated?

A complex line integral is calculated by breaking down the curve into small segments and approximating the work done by each segment. The sum of all these approximations gives the total work done along the curve.

What is the difference between a complex line integral and a regular line integral?

A regular line integral is calculated in the real plane, while a complex line integral is calculated in the complex plane. Additionally, a complex line integral involves the use of complex numbers and complex functions, while a regular line integral only involves real numbers and real functions.

What are some applications of complex line integrals?

Complex line integrals have various applications in physics and engineering, such as in the calculation of electric potential, magnetic flux, and fluid flow. They are also used in the study of complex analysis and in the development of mathematical models for physical systems.

Are there any special properties of complex line integrals?

Yes, there are several special properties of complex line integrals, such as the linearity property, the Cauchy integral theorem, and the Cauchy integral formula. These properties allow for efficient and accurate calculations of complex line integrals in various scenarios.

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