Calculate this integral: $$\oint_{\vert z\vert=1}\frac{e^{z^2+\sin(z)}}{4(z-2)^2e^{\cos(z)}}dz$$

It's obviously that I have to use the Cauchy Integral Theorem respectively formula. I read the solution and there they said, that the this integrand is holomoprhic in a surrounding $\bar{B}(0,1)$ so with CIT there one get that the value of this integral is 0.

I'm wondering why for example this integral: $$\oint_{\vert z\vert =2}\frac{\cos(z)}{z^2+4z+3}dz$$ one can't use the CIT and only the CIF. Thanks for your help!

  • $\begingroup$ The second integral isn't analytic in $z=-1$. $\endgroup$
    – Nosrati
    Commented Aug 17, 2017 at 10:09
  • $\begingroup$ @MyGlasses Ah, I see. So I have to use the CIF if there is a point, where the function isn't analytic? $\endgroup$
    – jacmeird
    Commented Aug 17, 2017 at 10:10
  • 1
    $\begingroup$ yes............ $\endgroup$
    – Nosrati
    Commented Aug 17, 2017 at 10:11

1 Answer 1


CIT= Cauchy Integral's Theorem stand for holomorphic function (without polar or essential singularity) on a domain $\Omega \subset \mathbb C$. Therefore for any curve $\gamma \subset\subset \Omega $ such that a function $f$ is holomorphic on $\Omega$. we have $$\oint_{\gamma} f(z)dz = 0$$. Which is the case for the function $$ f : D(0,1+\epsilon)\to \mathbb C, z\mapsto \frac{e^{z^2+\sin(z)}}{4(z-2)^2e^{\cos(z)}} $$. $0<\epsilon <<1$ is small enough. and $\gamma = \mathcal{C}(0,1) =\partial D(0,1)$. $f$ has no singularities in $\Omega = D(0,1+\epsilon)$

But CIF = Cauchy Integral's Formula applies for meromorphic functions on $\Omega $ (means holomorphic except at some locally finite singularities.) Hence one needs to use Residues theorem.

let $a \in \overset{\circ}{\gamma}$ where $\gamma \subset \subset \Omega $ is curve in the domain omega. Let $f: \Omega \to \mathbb C$ be holomorphic then the function $$ g: z\mapsto \frac{f(z)}{(z-a)^{n+1}}$$ has only one singularity (pole) at $z= a$ The CIF infer that,

$$\oint_{\gamma} g(z)dz = 2\mathrm{i}\pi Res(g,a) = 2\mathrm{i}\pi f^{n}(a)/n! $$

that is $$ f^{n}(a)= \frac{n!}{2\mathrm{i}\pi}\oint_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz. $$ Which is applicable to the second case where $$ \frac{\cos(z)}{z^2+4z+3}=\frac{\cos(z)}{(z-1)(z-3)} = \frac{f(z)}{z-1}$$

with $f(z) = \frac{\cos(z)}{z-3}$ is holomorphic on $D(0,2 +\frac{1}{2})$ without any singularities therein. And $\gamma = \mathcal{C}(0,2)= \partial D(0,2)\subset \subset D(0,3/2).$ So in this case we have

$$ \oint_{\gamma} g\frac{\cos(z)}{z^2+4z+3}dz = 2i\pi f(1) = -i\pi \cos(1)$$


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