Question 3.5: Prove that a (a) necessary and (b) sufficient condition that...
Prove that a (a) necessary and (b) sufficient condition that w = f(z) = u(x, y) + iv(x, y) be analytic in a region R is that the Cauchy –Riemann equations ∂u/∂x = ∂v/∂y, ∂u/∂y = -(∂v/∂x) are satisfied in R where it is supposed that these partial derivatives are continuous in R.
(a) Necessity. In order for f(z) to be analytic, the limit
\begin{aligned} \lim _{\Delta z \rightarrow 0} \frac{f(z+\Delta z)-f(z)}{\Delta z} &=f^{\prime}(z) \\ &=\lim _{\substack{\Delta x \rightarrow 0 \\ \Delta y \rightarrow 0}} \frac{\{u(x+\Delta x, y+\Delta y)+i v(x+\Delta x, y+\Delta y)\}-\{u(x, y)+i v(x, y)\}}{\Delta x+i \Delta y} &(1) \end{aligned}must exist independent of the manner in which \Delta z (or \Delta x and \Delta y ) approaches zero. We consider two possible approaches.
Case 1. \Delta y=0, \Delta x \rightarrow 0. In this case, (1) becomes
\lim _{\Delta x \rightarrow 0}\left\{\frac{u(x+\Delta x, y)-u(x, y)}{\Delta x}+i\left[\frac{v(x+\Delta x, y)-v(x, y)}{\Delta x}\right]\right\}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}provided the partial derivatives exist.
Case 2. \Delta x=0, \Delta y \rightarrow 0. In this case, (1) becomes
Now f(z) cannot possibly be analytic unless these two limits are identical. Thus, a necessary condition that f(z) be analytic is
\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=-i \frac{\partial u}{\partial y}+\frac{\partial v}{\partial y} \quad \text { or } \quad \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}(b) Sufficiency. Since ∂u/∂x and ∂u/∂y are supposed to be continuous, we have
\begin{aligned} \Delta u &=u(x+\Delta x, y+\Delta y)-u(x, y) \\ &=\{u(x+\Delta x, y+\Delta y)-u(x, y+\Delta y)\}+\{u(x, y+\Delta y)-u(x, y)\} \\ &=\left(\frac{\partial u}{\partial x}+\epsilon_1\right) \Delta x+\left(\frac{\partial u}{\partial y}+\eta_1\right) \Delta y=\frac{\partial u}{\partial x} \Delta x+\frac{\partial u}{\partial y} \Delta y+\epsilon_1 \Delta x+\eta_1 \Delta y \end{aligned}where \epsilon_1 \rightarrow 0 and \eta_1 \rightarrow 0 as \Delta x \rightarrow 0 and \Delta y \rightarrow 0.
Similarly, since partial v / \partial x and \partial v / \partial y are supposed to be continuous, we have
where \epsilon_2 \rightarrow 0 and \eta_2 \rightarrow 0 as \Delta x \rightarrow 0 and \Delta y \rightarrow 0. Then
\Delta w=\Delta u+i \Delta v=\left(\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}\right) \Delta x+\left(\frac{\partial u}{\partial y}+i \frac{\partial v}{\partial y}\right) \Delta y+\epsilon \Delta x+\eta \Delta y (2)
where \epsilon=\epsilon_1+i \epsilon_2 \rightarrow 0 and \eta=\eta_1+i \eta_2 \rightarrow 0 as \Delta x \rightarrow 0 and \Delta y \rightarrow 0
By the Cauchy-Riemann equations, (2) can be written
Then, on dividing by \Delta z=\Delta x+i \Delta y and taking the limit as \Delta z \rightarrow 0, we see that
\frac{d w}{d z}=f^{\prime}(z)=\lim _{\Delta z \rightarrow 0} \frac{\Delta w}{\Delta z}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}so that the derivative exists and is unique, i.e., f(z) is analytic in R.