1. ## Proof Cauchy-Goursat

Cauchy-Goursat Theorem: If $\displaystyle f(z)$ is analytic inside and on a simple, closed piecewise smooth curve $\displaystyle C$, then $\displaystyle \oint_{C} f(z) \ dz = 0$.

In proving the "special case" where we use Greens Theorem and the Cauchy Riemann Equations, is assuming that $\displaystyle f$ is continuously differentiable the same thing as saying that $\displaystyle f$ has continuous partial derivatives?

2. Originally Posted by manjohn12
Cauchy-Goursat Theorem: If $\displaystyle f(z)$ is analytic inside and on a simple, closed piecewise smooth curve $\displaystyle C$, then $\displaystyle \oint_{C} f(z) \ dz = 0$.

In proving the "special case" where we use Greens Theorem and the Cauchy Riemann Equations, is assuming that $\displaystyle f$ is continuously differentiable the same thing as saying that $\displaystyle f$ has continuous partial derivatives?
If $\displaystyle f$ is continous differenciable (on some open set) it means $\displaystyle f$ is differenciable and $\displaystyle f'$ is continous (on this open set). Write $\displaystyle f = g + hi$. Remember that $\displaystyle f ' = g_x + h_x i$. Thus, $\displaystyle g_x,h_x$ are continous on this open set. However, by Cauchy-Riemann equations $\displaystyle g_x=h_y$ and $\displaystyle h_x = -g_y$ so $\displaystyle h_y,g_y$ are continous also. Thus, the partial derivatives must be continous.

3. Originally Posted by ThePerfectHacker
If $\displaystyle f$ is continous differenciable (on some open set) it means $\displaystyle f$ is differenciable and $\displaystyle f'$ is continous (on this open set). Write $\displaystyle f = g + hi$. Remember that $\displaystyle f ' = g_x + h_x i$. Thus, $\displaystyle g_x,h_x$ are continous on this open set. However, by Cauchy-Riemann equations $\displaystyle g_x=h_y$ and $\displaystyle h_x = -g_y$ so $\displaystyle h_y,g_y$ are continous also. Thus, the partial derivatives must be continous.
You mean $\displaystyle f_x = g_x +ih_x$?

4. Originally Posted by manjohn12
You mean $\displaystyle f_x = g_x +ih_x$?
No. The derivative of $\displaystyle f$ is computed to be $\displaystyle g_x + ih_x$.