Contour Integral for Triangle with Non-Analytic Integrands

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SUMMARY

The discussion focuses on evaluating the contour integral \(\int \frac{z+1}{z^2 + 1} dz\) over a triangle with vertices at -3, 2i, and 3. The integrand is non-analytic at \(z^2 = \pm i\), but only the pole at \(i\) lies within the triangle. The solution is rewritten as \(\int \frac{\frac{z+1}{z+i}}{z-i} dz\) and evaluated using the residue theorem, yielding \(2 \pi i \cdot f(i) = 2 \pi i \cdot \frac{i+1}{2i}\). A minor correction regarding the notation of \(f(z_0)\) was noted, but the overall approach is validated.

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Homework Statement



let t be the triangle with vertices at the points -3, 2i, and 3, oriented counterclockwise. compute[tex]\int \frac {z+1}{z^2 + 1} dz[/tex]

Homework Equations



[tex]f(z) = \frac {1}{2 \pi i} * \int \frac {f(z)}{z - z_o} dz[/tex]

The Attempt at a Solution



the integrand fails to be analytic at z^2 = +/- i , but only the point i is inside the triangle t so I rewrote the equation as:

[tex]\int \frac {\frac {z+1}{z+i}}{z-i} dz[/tex]

= [tex]2 \pi i * f(i)[/tex]

= [tex]2 \pi i * \frac {i+1}{2i}[/tex]

is this correct? thanks!
 
Last edited:
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Good job! It looks good to me, but be sure to simplify your answer. Also, there was just a small typo in the relevant equations as the f(z) to the left of the equals sign should be f(z0), but this was obviously just a typo.
 

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