MHB Is My Complex Integral Calculation Correct?

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The integral calculation of $$ \int_\gamma \frac{z^3}{z-3} dz$$ over a circle of radius 4 centered at the origin is evaluated using the Cauchy integral formula. The solution confirms that the integral equals $$54\pi i$$. Multiple participants in the discussion affirm the correctness of this solution. The use of the Cauchy integral formula is appropriately applied in this context. Overall, the calculation is validated as accurate.
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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