We can consider complex valued functions as a type of vector fields.
\(\int_C f(r) ds=\lim_{\Delta s rightarrow 0 }\sum_{i=0}^n f(r(t_i))\Delta s_i\)
\(\int_C f(r) ds=\lim_{\Delta s rightarrow 0 }\sum_{i=0}^n f(r(t_i))\dfrac{\delta r(t_i)}{\delta t}\delta r_i\)
\(\int_C f(z) dz=\int_a^b f(r(t_i))\dfrac{\delta r(t_i)}{\delta t}\delta r_i\)
Previously we had partial differentiation on the real line. We could use the partial differention operator
We want to find a similar operator for the complex plane.
\(\int_C f(r) ds=\lim_{\Delta s rightarrow 0 }\sum_{i=0}^n f(r(t_i))\Delta s_i\)
\(\int_C f(r) ds=\lim_{\Delta s rightarrow 0 }\sum_{i=0}^n f(r(t_i))\dfrac{\delta r(t_i)}{\delta t}\delta r_i\)
\(\int_C f(z) dz=\int_a^b f(r(t_i))\dfrac{\delta r(t_i)}{\delta t}\delta r_i\)
If a function is complex differentiable, it is smooth.
Consider complex number z=x+iy
A function on this gives:
\(f(z)=u+iv\)
Take the total differential of :
\(df/dz=\dfrac{\delta f}{\delta z}+\dfrac{\delta f}{\delta x}\dfrac{dx}{dz}+\dfrac{\delta f}{\delta y}\dfrac{dy}{dz}\)
We know that:
\(\dfrac{dx}{dz}=1\)
\(\dfrac{dy}{dz}=-i\)
We can see from this that
\(\dfrac{du}{dx}=\dfrac{dv}{dy}\)
\(\dfrac{du}{dy}=-\dfrac{dv}{dx}\)
These are the Cauchy-Riemann equations