# How would Cauchy calculate $\int_{\partial B_1(2i)}\frac{e^{z^2}}{2i-z}dz$?

Please consider the following curve integral: $$I:=\int_{\partial B_1(2i)}\frac{e^{z^2}}{2i-z}dz$$ where $$B_r(z_0):=\left\{z\in\mathbb{C}:|z-z_0|<r\right\}$$

Let $\gamma :[a,b]\to\Omega$ denote a piecewise continuously differentiable path in $\Omega$, $\gamma^*:=\gamma([a,b])$ denote its trace, and $f:\gamma^*\to\mathbb{C}$ be continuously $\Rightarrow$ $$\int_\gamma f(z)dz:=\int_a^b\gamma '(t)f(\gamma (t))\;dt$$ is called curve integral of $f$ along $\gamma$.

Using the definition above, we can calculate $I$. However, I'm sure there is an easier way to calculate the integral. I thought about Cauchy's integral theorem or Cauchy's integral formula . I'm new to them and still unsure how they can be applied here.

• what is $B_1(2i)$?? ball centered at $2i$?? – Santosh Linkha Jun 2 '14 at 11:43
• When I tried using Cauchy's integral formula, I didn't get zero. Are you sure $I = 0$? – Michael Albanese Jun 2 '14 at 11:46
• The integral is not $0$ as can readily be seen by Cauchy's integral formula. – Git Gud Jun 2 '14 at 11:47
• @SantoshLinkha Sorry for that. I've provided the definition. – 0xbadf00d Jun 2 '14 at 12:11
• @GitGud I'm sorry if I'd made a mistake. But the main question is how Cauchy's integral theorem/formula can help here. I've updated my question accordingly. – 0xbadf00d Jun 2 '14 at 12:13

To use Cauchy's integral formula, in the notation of the wikipedia link you provided, set $f(z)=-e^{z^2}$ and $a=2i$. This gives you $I=-\left.2\pi ie^{z^2}\right|_{z=2i}=-2\pi e^{-4}i$.
As an added bonus, one can even infer that $z\mapsto -\dfrac{e^{z^2}}{z-2i}$ doesn't have an antiderivative in $\mathbb C$ (or in any neighborhood of $2i$) for if it did, since the path is closed, the integral would be $0$.
• If $$f(z):=\frac{e^{z^2}}{2i-z}$$ can't we restrict $f$'s domain to $B_1(2i)$? Then, since $B_1(2i)$ is star-shaped, Cauchy's integral theorem would be applicable. – 0xbadf00d Jun 2 '14 at 12:26
• @oxbadfood You would need $f$ holomorphic in $B_1(2i)$, it isn't even defined there. – Git Gud Jun 2 '14 at 12:27
• You're absolutely right. As stated in the comments above, I thought about $B_2(0)$ while writing $B_1(2i)$. Thanks a lot. – 0xbadf00d Jun 2 '14 at 12:32
• Cauchy's integral theorem doesn't require the domain to be simply connected. It requires that the cycle over which one integrates is null-homologous (the winding number around any point not in the domain is $0$). – Daniel Fischer Jun 2 '14 at 12:41