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Proof of Cauchy - Riemann equations

Theorem :

The necessary conditions that a function  f(z)=U(x, y) + i V(x, y) be analytic in a domain D is that U and V satisfy the Cauchy- Riemann equations i.e.\(\frac{\partial U}{\partial x}=\frac{\partial V}{\partial y};\frac{\partial U}{\partial y}=-\frac{\partial V}{\partial x}\).

Proof :

     Let f(z) is analytic in D, then f'(z) exists uniquely at every point of D . Therefore for all z \(\epsilon\) D

    \({f}'(z)=\lim_{\Delta z\rightarrow 0} \frac{f(x+\Delta)-f(z)}{\Delta z}\)  

exists, and is unique as \(\Delta z\rightarrow 0\) along any path we choose.

Now we consider the following two cases:

Case I. Suppose first that in

               \(\frac{f(z+\Delta z)-f(z)}{\Delta z}\)

\(\Delta z\) approaches zero along the real axis or x-axis then \(\Delta z = \Delta x\) and \(\Delta y = 0\).

        Thus \(\frac{(z + \Delta z) - f(z)}{\Delta z}\)

                 = \(\frac{u(x +\Delta x, y) + iv(x + \Delta x, y) - u(x,y) - iv(x,y)}{\Delta x}\)

                 = \(\frac{u(x + \Delta x,y) - u(x,y)}{\Delta x}\) + i \(\frac{v(x +\Delta x,y)-v(x,y)}{\Delta x}\)

        Now

                 \(\lim_{\Delta z \rightarrow 0} \frac{f(z+\Delta z)-f(z)}{\Delta z}\)

                 = \(\lim_{\Delta x \rightarrow 0} \frac{u(x,y + \Delta y) - u(x,y)}{\Delta x}\) + i  \(\lim_{\Delta x \rightarrow 0} \frac{v(x + \Delta x,y) - v(x,y)}{\Delta x}\)

       Thus  f'(z) = \(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}\)  .........(1)

Case II. Next let \(\Delta z\) approaches zero along the imaginary axis (or y-axis), then \(\Delta z\) = i\(\Delta y, \Delta x = 0\) and we have

             \( \frac{f(z + \Delta z) - f(z)}{\Delta z}\)

             = \(\frac{u(x, y+\Delta y) + iv(x, y+\Delta y) - u(x,y) - iv(x,y)}{i\Delta y}\)

             = \(\frac{u(x, y+ \Delta y) - u(x,y)}{i\Delta y}\) + \(\frac{v(x, y+\Delta y)-v(x,y)}{\Delta y}\)

       Thus

       f'(z) = \(\lim_{\Delta z \rightarrow 0} \frac{f(z+\Delta z)-f(z)}{\Delta z}\)

               = \(\lim_{\Delta y \rightarrow 0} \frac{u(x,y + \Delta y) - u(x,y)}{i\Delta y}\) +   \(\lim_{\Delta y \rightarrow 0} \frac{v(x, y+ \Delta y) - v(x,y)}{\Delta y}\)

               = \( \frac{1}{i} \frac{\partial u}{\partial y} + \frac(\partial v}{\partial y} \)

        \( \therefore  f'(z) \)= \( \frac{\partial v}{\partial y}  - i \frac{\partial u}{\partial y) \)   .........(2)

From (1) and (2), we have

              f'(z) = \(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}

                      = \(\frac{\partial v}{\partial y} + i \frac{\partial u}{\partial y}

Equating real and imaginary parts in above, we obtain

                      \(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial v}{\partial x} = \frac{-\partial u}{\partial y}  ............(3)

Equations (3), known as the  Cauchy-Riemann equations give the necessary condition for a function f(z) to be analytic.


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