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Good evening everyone,

I want to know if my result is correct. So:

I have to compute the following integral:

$$\int_\gamma \frac{ze^{\pi z}}{z^2+1}dz,$$ while $\gamma:[0,3]\to\mathbb{C},\gamma(t)=-3^{2i\pi t}$.

Firstly, I've found the pols of first order $z_1=+i, z_2=-i$. Then, by Cauchy's residue theorem, is that $$\int_\gamma f(z)dz = 2\pi i\sum_{k=1}^n \text{Res}_{a_k}f\cdot n(\gamma,a_k).$$ So, my residues are: $$\text{Res}_if = \lim_{z\to i}(z-i)f(z) =\left.\frac{ze^{\pi z}(z-i)}{(z+i)(z-i)} \right|_{z=i}=\frac{ie^{i\pi}}{2i}=-\frac{1}{2},$$ $$\text{Res}_{-i}f = \lim_{z\to -i}(z+i)f(z) =\left.\frac{ze^{\pi z}(z+i)}{(z+i)(z-i)} \right|_{z=-i}=\frac{-ie^{i\pi}}{-2i}=-\frac{1}{2}.$$ Now the radius is $r=3$, so the singularities are in the sphere $B(0,3)$, so their winding numbers are 3 because of $\gamma$. All in all: \begin{align} \int_\gamma \frac{ze^{\pi z}}{z^2+1}dz& =2\pi i\sum_{k=1}^2n(\gamma,z_k)\cdot\text{Res}_{z_k}f \\ &=2\pi i\left[-\frac{1}{2}n(\gamma,i)-\frac{1}{2}n(\gamma,-i) \right]\\ &=-6\pi i. \end{align} Is it right? Please help me :)

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  • $\begingroup$ Your definition of $\gamma$ seems wrong. $\gamma(t) = -3^{2i\pi t} = -e^{2i\pi t\log 3}$ doesn't give a closed curve for the parameter interval $[0,3]$. Should it have been $\gamma(t) = -3 e^{2i\pi t}$? $\endgroup$ Commented May 28, 2015 at 18:15
  • $\begingroup$ The parameter for this integral should be $\gamma(t)=-3e^{2i\pi t}$, yes. $\endgroup$
    – MathCracky
    Commented May 28, 2015 at 18:30

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After fixing the definition of $\gamma$ (presumably a typo) to $\gamma(t)= -3e^{2i\pi t}$, there is one slip-up in the computation of the residue at $-i$ (which could also be just a typo):

$$\operatorname{Res}_{-i} f = \frac{-ie^{-i\pi}}{-2i} = \frac{1}{2} e^{-i\pi}.$$

As it so happens, we have $e^{i\pi} = e^{-i\pi}$, but still the correct thing is to write $e^{-i\pi}$ there.

Then we have a confusing sentence:

Now the radius is $r=3$, so the singularities are in the sphere $B(0,3)$, so their winding numbers are $3$ because of $\gamma$.

Split that up, "Now the radius is $r =3$, so the singularities are in the disk $B(0,3)$. [I use disk instead of "sphere", since in the terminology I grew up with a sphere is the boundary of a ball/disk.] The winding number around both is $3$, since $\gamma$ traverses the circle thrice counterclockwise."

Or find a better formulation.

But, most importantly, your arguments are correct, just one minor slip-up, and a less-than-optimal formulation.

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