Complex Integral Homework: Calculate (z+(1/z))^n dz

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SUMMARY

The integral of (z+(1/z))^n dz over the unit circle C can be approached using the substitution z=e^(i*theta). This transforms the integral into i ∫_0^(2π) (2*i*sin(theta))^n e^(i*theta) dtheta. The discussion highlights the need to handle two cases based on whether n is odd or even, as this affects the integration of the resulting expressions involving cosines and sines. The final form involves terms like (2*cos(theta))^n * e^(i*theta), which require careful expansion and integration techniques to solve.

PREREQUISITES
  • Complex analysis fundamentals
  • Knowledge of contour integration
  • Familiarity with Euler's formula (e^(i*theta) = cos(theta) + i*sin(theta))
  • Understanding of odd and even functions in integration
NEXT STEPS
  • Study contour integration techniques in complex analysis
  • Learn about the residue theorem for evaluating complex integrals
  • Explore the properties of odd and even functions in calculus
  • Investigate the use of Fourier series in relation to integrals of periodic functions
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Students of complex analysis, mathematicians tackling integral calculus, and anyone interested in advanced techniques for evaluating complex integrals.

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Homework Statement


Homework Equations




I hope there's someone who can help me with the following:

I have to calculate the integral over C (the unit cicle) of (z+(1/z))^n dz, where z is a complex number.

The Attempt at a Solution




I tried to use the subtitution z=e^(i*theta), so you get
(z+(1/z))^n dz=(2*i*Sin(theta))^n * i*e^(i*theta) dtheta
but then I get stuck.
Is this the right way, and if, how do I proceed. And if it isn't, how should I do it?
 
Last edited:
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i suppose you integral looks like:
\int_{|z|=1} (z+1/z)^n dz = i \int_0^{2\pi} (e^{i\theta}+e^{-i\theta})^n e^{i\theta} d\theta

now did what you did then also try to expand the remaining e^{i\theta}=\cos \theta +i \sin \theta, and now you end up with two integrals with just cos and sin... you can then do the integral for two cases n odd and n even... etc...
 
Ok, but then you get:

(2*cos(theta))^n *e^(i*theta)

but I don't know how to get rid of the n...

(Don't know to use latex...)
 
i said to use e^{i\theta}=\cos \theta +i \sin \theta to expand the second exponential.. and then multiply out to get something like
\cos^{n+1} \theta + \cos^n \theta \sin \theta and now you can try integrate these assuming that n is an integer. I am guessing that there will be two cases: n odd an n even
 

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