Complex Analysis: Cauchy's Theorem

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Cauchy's Theorem states that if functions u and v satisfy the Cauchy-Riemann equations within and on a simple closed contour C, then the integral of the function f(z) equals zero. In the example provided, with f(z) = 1/(z+20) and a closed contour as a circle of radius 1 around the origin, the integral is indeed zero because the singularity at z = -20 lies outside the contour. This confirms that f is differentiable within and on the circle |z|=1. Understanding these concepts is crucial for exam preparation. Mastery of Cauchy's Theorem is essential for complex analysis.
jokusan13
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As I am studying for an exam I am trying to wrap my head around the concepts I learned. I want to make sure I fully understand the concepts before the exam in 1.5 weeks.

Cauchy's Theorem
If u and v satisfy the Cauchy-Riemann equations inside and on the simple closed contour C, then the integral of f(z)=0

Now for example, if we have f(z)=1/(z+20) and our closed contour is a circle around the origin with radius=1. If I am understanding this correctly, we can say that the integral is equal to 0 since the 'bad point' of z=-20 is outside of the circle correct meaning that f is differentiable in and on |z|=1.

Does this sound correct?
 
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Yes that sounds correct.
 

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