MHB Is My Complex Integral Calculation Correct?

hmmmmm
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Is my solution to the following problem correct?

Evaluate $$ \int_\gamma \frac{z^3}{z-3} dz$$ where $\gamma$ is the circle of radius 4 centered at the origin.

Solution

Form the cauchy integral formula we have that:

$$ f(3)=\frac{1}{2\pi i} \int_\gamma \frac{z^3}{z-3}dz$$ and so $$ \int_\gamma \frac{z^3}{z-3} dz=54\pi i $$Thanks very much for any help
 
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hmmm16 said:
Is my solution to the following problem correct?

Evaluate $$ \int_\gamma \frac{z^3}{z-3} dz$$ where $\gamma$ is the circle of radius 4 centered at the origin.

Solution

Form the cauchy integral formula we have that:

$$ f(3)=\frac{1}{2\pi i} \int_\gamma \frac{z^3}{z-3}dz$$ and so $$ \int_\gamma \frac{z^3}{z-3} dz=54\pi i $$Thanks very much for any help

Looks fine.
 
hmmm16 said:
Is my solution to the following problem correct?

Evaluate $$ \int_\gamma \frac{z^3}{z-3} dz$$ where $\gamma$ is the circle of radius 4 centered at the origin.

Solution

Form the cauchy integral formula we have that:

$$ f(3)=\frac{1}{2\pi i} \int_\gamma \frac{z^3}{z-3}dz$$ and so $$ \int_\gamma \frac{z^3}{z-3} dz=54\pi i $$Thanks very much for any help

Hi hmmm16,

Your answer is correct.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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