Polar Cauchy–Riemann Equations and Laplace Equation Explained Step by Step

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This post presents a complete and detailed derivation of the polar form of the Cauchy–Riemann equations and the corresponding Laplace equation in polar coordinates. Starting from the definition of the complex derivative, each step is developed rigorously by considering radial and angular increments of the complex variable. No intermediate steps are omitted, making the exposition especially useful for postgraduate students, research scholars, and aspirants preparing for competitive examinations in Complex Analysis.

Polar Form of the Cauchy–Riemann Equations and Laplace Equation

Let \[ f(z) = u(r,\theta) + i v(r,\theta) \] be differentiable at a point \(z\), where \[ z = r e^{i\theta}. \]

By definition, the derivative of \(f\) at \(z\) is \[ f'(z) = \lim_{h \to 0} \frac{f(z+h) - f(z)}{h}. \]

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Step 1: Radial Increment

Choose nearby points of \(z\) along the radial direction. Let \[ z + h = (r + \Delta r)e^{i\theta}. \] Then \[ h = \Delta r \, e^{i\theta}, \; \Delta r \neq 0. \]

Now compute \[ \frac{f(z+h) - f(z)}{h} \]

\[= \frac{u(r+\Delta r,\theta) + i v(r+\Delta r,\theta) - u(r,\theta) - i v(r,\theta)}{\Delta r e^{i\theta}}. \]

Separating real and imaginary parts: \[ = \frac{1}{e^{i\theta}} \bigg[ \frac{u(r+\Delta r,\theta)-u(r,\theta)}{\Delta r} \] \[+ i \frac{v(r+\Delta r,\theta)-v(r,\theta)}{\Delta r} \bigg]. \]

Letting \(\Delta r \to 0\), we obtain \[ f'(z) = \frac{1}{e^{i\theta}} \left( u_r + i v_r \right). \tag{1.01} \]

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Step 2: Circular Increment

Now choose nearby points of \(z\) along the circular arc through \(z\). Let \[ z + h = r e^{i(\theta + \Delta\theta)} = r e^{i\theta} e^{i\Delta\theta}. \] Thus \[ h = r e^{i\theta}(e^{i\Delta\theta} - 1), \; \Delta\theta \neq 0. \]

Now compute \[ \frac{f(z+h) - f(z)}{h} = \]

\[ \frac{u(r,\theta+\Delta\theta) + i v(r,\theta+\Delta\theta) - u(r,\theta) - i v(r,\theta)} {r e^{i\theta}(e^{i\Delta\theta}-1)}. \]

Rewrite as \[ = \frac{1}{i r e^{i\theta}} \bigg[ \frac{u(r,\theta+\Delta\theta)-u(r,\theta)}{\Delta\theta} +\] \[ i \frac{v(r,\theta+\Delta\theta)-v(r,\theta)}{\Delta\theta} \bigg] \cdot \frac{i\Delta\theta}{e^{i\Delta\theta}-1}. \]

Since \[ \lim_{\Delta\theta\to 0} \frac{i\Delta\theta}{e^{i\Delta\theta}-1} = 1, \] letting \(\Delta\theta \to 0\) gives \[ f'(z) = \displaystyle\frac{1}{i r e^{i\theta}} (u_\theta + i v_\theta)\] \[= \displaystyle \frac{1}{r e^{i\theta}} (v_\theta - i u_\theta). \tag{1.02} \]

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Step 3: Polar Cauchy–Riemann Equations

Since expressions (1.01) and (1.02) represent the same derivative \(f'(z)\), we equate them: \[ u_r + i v_r = \frac{1}{r}(v_\theta - i u_\theta). \]

Equating real and imaginary parts yields \[ u_r = \frac{1}{r} v_\theta, \; v_r = -\frac{1}{r} u_\theta. \tag{1.03} \]

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Step 4: Derivation of Laplace Equation (Polar Form)

Differentiate the first equation of (1.03) with respect to \(r\): \[ u_{rr} = \frac{1}{r} v_{\theta r} - \frac{1}{r^2} v_\theta. \]

Differentiate the second equation of (1.03) with respect to \(\theta\): \[ v_{r\theta} = -\frac{1}{r} u_{\theta\theta}. \]

Using equality of mixed partial derivatives: \[ v_{\theta r} = v_{r\theta}, \] we substitute and obtain \[ u_{rr} = -\frac{1}{r^2} v_\theta - \frac{1}{r} u_{\theta\theta}. \]

Since \(v_\theta = r u_r\), we finally get \[ u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} = 0. \]

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Special Case

If \(u(r,\theta)\) depends only on \(r\), then \[ u_{rr} + \frac{1}{r}u_r = 0. \]

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