MHB Finding the Primitive of a Complex Integral

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How can I find the primitive of $\int_{\gamma}ze^{z^2}dz$ from $i$ to $2-i$?
 
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dwsmith said:
How can I find the primitive of $\int_{\gamma}ze^{z^2}dz$ from $i$ to $2-i$?

$$\int z e^{z^{2}}\,dz=\frac{1}{2}\int 2z e^{z^{2}}\,dz.$$

Can you finish?
 
Ackbach said:
$$\int z e^{z^{2}}\,dz=\frac{1}{2}\int 2z e^{z^{2}}\,dz.$$

Can you finish?

So $\left(\frac{e^{z^2}}{2}\right)'=\int ze^{z^2}dz$ Then to solve the integral I just integrate g'(z) right?
 
dwsmith said:
So $\left(\frac{e^{z^2}}{2}\right)'=\int ze^{z^2}dz$ Then to solve the integral I just integrate g'(z) right?

Actually, I would have said that

$$\left(\frac{e^{z^{2}}}{2}\right)'=ze^{z^{2}}.$$

Then just use the Fundamenal Theorem of the Calculus, which works because your function is analytic.
 
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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