I've been working with Complex Analysis and have noticed an interesting result.(adsbygoogle = window.adsbygoogle || []).push({});

Under what conditions will the following integral be purely imaginary:

[tex]\int_{a-bi}^{a+bi} f(z)dz[/tex]

It seems to me some type of symmetry is required. Take for example:

[tex]\int_{1-8i}^{1+8i} f(z)dz[/tex]

where:

[tex]f(z)= \frac{e^{3z}}{z}[/tex]

Now:

[tex]\int_{1-8i}^{1+0i} \frac{e^{3z}}{z}dz\approx 10.6559+2.7271i[/tex]

and:

[tex]\int_{1+0i}^{1+8i} \frac{e^{3z}}{z}dz\approx -10.6559 + 2.7271i[/tex]

Thus:

[tex]\int_{1-8i}^{1+8i} \frac{e^{3z}}{z}dz\approx 5.4543i[/tex]

Note the plot below which is the image of f(z) in the u-v plane for f(z) along the path indicated above. Red is the image for negative values of b; blue is for positive values. The symmetry is obvious.

I've noticed this with other similar integrals and suspect symmetry causes the integral to be purely imaginary but haven't yet figured out how to prove it. Can anyone offer ideas/hints/suggestions?

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# Purely imaginary complex contour

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