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.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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