# Thread: Complex Anaylsis: NEED HELP

1. ## Complex Anaylsis: NEED HELP

I have 2 questions regarding complex analysis:

1) Let f be analytic inside and on a closed contour gamma. Show that

the integral of [f'(z)/(z-w)] dz = the integral of [f(z)/(z-w)^2] dz

for all w not on Gamma.

2) Explain why f(z) = e^z^2 has an antiderivative in the whole plane.

2. Originally Posted by Vitava61
1) Let f be analytic inside and on a closed contour gamma. Show that

the integral of [f'(z)/(z-w)] dz = the integral of [f(z)/(z-w)^2] dz
Since $\displaystyle f$ is analytic on a simply connected set containing $\displaystyle \Gamma$ it means,
$\displaystyle \frac{1}{2\pi i}\oint_{\Gamma}\frac{f(z)}{(z-w)^2} = f''(w)$
And that,
$\displaystyle \frac{1}{2\pi i}\oint_{\Gamma} \frac{f'(z)}{(z-w)} = [f'(z)]'|_w = f''(w)$
Both follow by the Cauchy Integral Formula.
2) Explain why f(z) = e^z^2 has an antiderivative in the whole plane.
Define $\displaystyle F(z) = \int_{[0,z]}e^{w^2} dw$.
Where $\displaystyle [0,z]$ is the line segment connecting $\displaystyle 0$ to $\displaystyle z$.
Then, $\displaystyle F'(z) = e^{z^2}$.