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This is the exercise $25$, Section $14.3$ (Cauchy's Integral Theorem) from Advanced Engineering Mathematics. It was translated to Portuguese and I do not have the book in English. So I will try to translate the exercise to English.

I want to evaluate the following integral $\oint_{C}\dfrac{\cos z}{z} dz$, where $C$ is the curve consisting of $|z|=1$ (anticlockwise) and $|z|=3$ (clockwise). Why $C$ is a curve?

I would appreciate your help.

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What do you mean by "$C$ is the curve consisting of $|z|=1$ and $|z|=3$?" –  Argon Sep 19 '12 at 1:17
    
@Argon: I do not know what it means. I tried to improve the question. –  spohreis Sep 19 '12 at 1:35
    
Is it perhaps two different curves that you are supposed to evaluate the integral on? –  Argon Sep 19 '12 at 1:37

1 Answer 1

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While I'm not sure what your book means by " a curve consisting of $|z|=1$ (anticlockwise) and $|z|=3$ (clockwise)", I believe this should clear things up for you as to why $|z| = C$ is a curve.

Recalling that if $z=x+iy$ where $x,y \in \mathbb {R}$ and $C \in \mathbb {R}^+$

$$|z|=\sqrt{x^2+y^2}=C$$

So if we let $z = C\cos \theta + i C\sin \theta = C e^{i \theta}$

$$|z|=\sqrt{C^2(\cos^2 \theta+\sin^2 \theta)} = |C| = C$$

So we see that the curve $|z|=C$ is a circle of radius $C$ centred at the origin of the complex plane, which can be parametrized by $z(\theta) = Ce^{i\theta}$.

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