Question

To prove that the Jacobian of the transformation is $\frac{\mathbf{\partial }\left(u,v\right)}{\mathbf{\partial }\left(x,y\right)}\mathbf{=}{\left|f\text{'}\left(z\right)\right|}^{\mathbf{2}}$ using Cauchy- Reimann equations.

Short answer

$J\left(\frac{u,v}{v}\right)={\left|f\text{'}\left(z\right)\right|}^2$

Step-by-step solution

Concept of Cauchy Riemann theorem

Formula from Cauchy Riemann theorem:

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$And $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$

Jacobian formula:

$J\left(\frac{u,v}{x,y}\right)=\left|\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial u}{\partial y}\end{array}\right|$

Use Cauchy Riemann theorem for calculation

Let, f(z) = u+ iv. ...... (1)

Differentiating equation (1) with respect to x is given as:

$f\text{'}\left(z\right)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}$ ...... (2)

Taking modulus of the equation (2) as follows:

$$\begin{gathered} \left|f\text{'}\left(z\right)\right|=\sqrt{{\left(\frac{\partial u}{\partial x}\right)}^2+{\left(\frac{\partial v}{\partial x}\right)}^2} \\ {\left|f\text{'}\left(z\right)\right|}^2={\left(\frac{\partial u}{\partial x}\right)}^2+{\left(\frac{\partial v}{\partial x}\right)}^2 \end{gathered}$$ ...... (3)

Jacobian of f(z) is given as follows:

$J\left(\frac{u,v}{x,y}\right)=\left|\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right|$ ....... (4)

Cauchy Riemann- condition is given as follows:

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$And $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$. ...... (5)

Put equation (4) in (5)

$\begin{array}{rcl}J\left(\frac{u,v}{x,y}\right)& =& \left|\begin{array}{cc}\frac{\partial u}{\partial x}& -\frac{\partial v}{\partial x}\\ \frac{\partial v}{\partial x}& \frac{\partial u}{\partial x}\end{array}\right|\\ & \Rightarrow & \left(\frac{\partial u}{\partial x}·\frac{\partial u}{\partial x}\right)-\left(-\frac{\partial v}{\partial x}·\frac{\partial v}{\partial x}\right)\\ & \Rightarrow & {\left(\frac{\partial u}{\partial x}\right)}^2+{\left(\frac{\partial v}{\partial x}\right)}^2\\ J\left(\frac{u,v}{x,y}\right)& =& {\left|f\text{'}\left(z\right)\right|}^2\\ & & \end{array}$

Hence, $\begin{array}{rcl}J\left(\frac{u,v}{x,y}\right)& =& {\left|f\text{'}\left(z\right)\right|}^2\\ & & \end{array}$.