M4.2 Cauchy Riemannian Equations in Cartesian Form/Polar Form

Cauchy Riemannian Equations in Cartesian Form - (VTU 2020,2019,2018,2017)

The necessary condition for a complex function \(f(z)=u(x,y)+i v(x,y)\) is said to be analytic is that, there exists four first order continuous partial derivatives \(\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}\) such that it satisfies the conditions $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}$$ $$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$$ Proof: Let \(f(z)=u(x,y)+iv(x,y)\) be a complex function and is analytic at \(z=x+iy\). According to definition of Analytic function we have \(f'(z)=lim_{\delta z\to z}{\frac{f(z+\delta z)-f(z)}{\delta z}}\). Let \(\delta z= \delta x+i \delta y\) then \(f(z+\delta z)= u(x+\delta x,y+\delta y)+i v(x+\delta x,y+\delta y)\) $$\Longrightarrow f’(z)=\frac{[u(x+\delta x,y+\delta y)-u(x,y)]+i[v(x+\delta x,y+\delta y)-v(x,y)]}{\delta x+ i \delta y}$$ $$\Longrightarrow f’(z)= lim_{\delta z\to 0}\frac{[u(x+\delta x,y+\delta y)-u(x,y)]}{ \delta x+ i \delta y }+i lim_{\delta z\to 0} \frac{[v(x+\delta x,y+\delta y)-v(x,y)]}{\delta x+ i \delta y}$$ Case 01: If \(z\) approaches horizontally, i.e., \(\delta y=0\) then \(\delta z = \delta x\) $$\Longrightarrow f’(z)= lim_{\delta x\to 0}\frac{[u(x+\delta x,y)-u(x,y)]}{\delta x}+i lim_{\delta x\to 0} \frac{[v(x+\delta x,y)-v(x,y)]}{\delta x}$$ $$\Longrightarrow f’(z)= \frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} \longrightarrow (1)$$
Case 02: If \(z\) approaches vertically, i.e., \(\delta x=0\) then \(\delta z = i\delta y\) $$\Longrightarrow f’(z)= lim_{i\delta y\to 0}\frac{[u(x,y+\delta y)-u(x,y)]}{i\delta y}+ lim_{i\delta y\to 0} \frac{[v(x,y+\delta y)-v(x,y)]}{\delta y}$$ $$\Longrightarrow f’(z)= -i\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}\longrightarrow (2)$$ Comparing (1) and (2) we will get $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}, \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$$

Cauchy Riemannian Equations in Polar Form - (VTU 2020,2019,2018,2017)

The necessary condition for a complex function \(f(z)=u(r,\theta)+i v(r,\theta)\) is said to be analytic is that, there exists four first order continuous partial derivatives \(\frac{\partial u}{\partial r}, \frac{\partial u}{\partial \theta}, \frac{\partial v}{\partial r}, \frac{\partial v}{\partial \theta}\) such that it satisfies the conditions $$\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta}$$ $$\frac{\partial v}{\partial r}=-\frac{1}{r}\frac{\partial u}{\partial \theta}$$ Proof:Let \(f(z)=u(r,\theta)+iv(r,\theta)\) be a complex function and is analytic at \(z=r e^{i\theta}\). $$\Longrightarrow f(z)=f(r e^{i\theta})= u(r,\theta)+iv(r,\theta)=u+iv$$ Since \(f\) is an Analytic function, \(f’(z)\) is exists and unique. Differentiate \(f\) with respect to \(r\) $$\Longrightarrow f’(z)=f’(r e^{i\theta}) e^{i\theta}= \frac{\partial u}{\partial r}+i\frac{\partial v}{\partial r} \longrightarrow (1)$$ Differentiate \(f\) with respect to \(\theta\) $$\Longrightarrow f’(z)=f’(r e^{i\theta})\cdot r\cdot e^{i\theta}\cdot i= \frac{\partial u}{\partial \theta}+i\frac{\partial v}{\partial \theta} \longrightarrow (2)$$ Put (1) in (2), we will get $$\Longrightarrow r\cdot i\left(\frac{\partial u}{\partial r}+i\frac{\partial v}{\partial r}\right)= \frac{\partial u}{\partial \theta}+i\frac{\partial v}{\partial \theta}$$ $$\Longrightarrow ir\frac{\partial u}{\partial r}-r\frac{\partial v}{\partial r}= \frac{\partial u}{\partial \theta}+i\frac{\partial v}{\partial \theta}$$ Comparing real and imaginary part of the above equation we get $$\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta}$$ $$\frac{\partial v}{\partial r}=-\frac{1}{r}\frac{\partial u}{\partial \theta}$$