Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\gamma(z_0,R)$ denote the circular contour $z_0+Re^{it}$ for $0\leq t\leq 2\pi$. Evaluate: $$\int_{\gamma(0,1)} \frac{\cos(z)}{z^2}dz$$

share|improve this question
    
What have you tried? Cauchy's integral formula or the residue theorem work well here. –  Ragib Zaman Dec 20 '12 at 14:39

2 Answers 2

${\cos(z)\over z^2}={1\over z^2}(1-{z^2\over 2!}+{z^4\over 4!}-...)={1\over z^2}-{1\over 2!}+{z^2\over 4!}-...\implies$ Coefficient of ${1\over z}$ in the Laurent series expansion of ${\cos(z)\over z^2}=0\implies$ $\int_{\gamma(0,1)} \frac{\cos(z)}{z^2}dz=2\pi i\times0=0.$

share|improve this answer

I don't know what you mean by $\gamma(0,1)$ but I will assume you want the closed line integral over $\gamma$. If $D=\left\{\left|z-z_0\right|<R\right\}$ then:

If $0\in D$, by Cauchy's differentiation formula, $$\oint_{\gamma}\frac{\cos z}{z^2}dz=2\pi i f^{\prime}(0)$$ where $f(z)=\cos z$

If $0\notin D$ then $0\notin \gamma([0,2\pi])$ (so that your integral is defined) and by Cauchy's Integral Theorem, $$\oint_{\gamma}\frac{\cos z}{z^2}dz=0$$ ($f$ is analytic in $D$)

share|improve this answer
1  
Shouldn't it be $f(z) = \cos z$? (and thus your distinction into cases isn't really necessary) –  Fabian Dec 20 '12 at 15:48
    
@Fabian Yeah I will edit my answer –  Nameless Dec 20 '12 at 16:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.