1. Find amplitude of \( \frac{3+2 i}{3-2 i} \) \( \frac{3+2 i}{3-2 i} \) का कोणांक ज्ञात कीजिए। 2. Show that \( \mathrm{f}(\mathrm{z})=\bar{z} \) is not differentiable at any point. सिद्ध कीजिए कि \( \mathrm{f}(\mathrm{z})=\bar{z} \) कसी भी बिन्दु पर अवकलनीय नहीं है। 3. Show that \( \mathrm{u}=\mathrm{x}^{3}-3 \mathrm{xy}^{2} \) is a harmonic function. 4. सिद्ध कीजिए \( \mathrm{u}=\mathrm{x}^{3}-3 \mathrm{xy}^{2} \) एक प्रंसवादी फलन है। Find the radius of convergence of the power series.

Let's solve each part of your problem step by step. ### 1. Find the amplitude of \( \frac{3+2 i}{3-2 i} \) To find the amplitude (or argument) of the complex number \( \frac{3+2 i}{3-2 i} \), we can first simplify it. **Step 1: Simplify the expression.** Multiply the numerator and denominator by the conjugate of the denominator: \[ \frac{3+2i}{3-2i} \cdot \frac{3+2i}{3+2i} = \frac{(3+2i)(3+2i)}{(3-2i)(3+2i)} \] **Step 2: Calculate the denominator.** \[ (3-2i)(3+2i) = 3^2 - (2i)^2 = 9 - (-4) = 9 + 4 = 13 \] **Step 3: Calculate the numerator.** \[ (3+2i)(3+2i) = 3^2 + 2 \cdot 3 \cdot 2i + (2i)^2 = 9 + 12i - 4 = 5 + 12i \] **Step 4: Combine results.** Now we have: \[ \frac{3+2i}{3-2i} = \frac{5 + 12i}{13} = \frac{5}{13} + \frac{12}{13}i \] **Step 5: Find the amplitude.** The amplitude (or argument) \( \theta \) can be found using: \[ \theta = \tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \tan^{-1}\left(\frac{12/13}{5/13}\right) = \tan^{-1}\left(\frac{12}{5}\right) \] Now, let's calculate this value. ### 2. Show that \( f(z) = \bar{z} \) is not differentiable at any point. To show that \( f(z) = \bar{z} \) is not differentiable, we can use the Cauchy-Riemann equations. **Step 1: Express \( z \) in terms of \( x \) and \( y \).** Let \( z = x + iy \), then \( \bar{z} = x - iy \). **Step 2: Identify \( u \) and \( v \).** Here, \( u(x, y) = x \) and \( v(x, y) = -y \). **Step 3: Calculate partial derivatives.** \[ \frac{\partial u}{\partial x} = 1, \quad \frac{\partial u}{\partial y} = 0, \quad \frac{\partial v}{\partial x} = 0, \quad \frac{\partial v}{\partial y} = -1 \] **Step 4: Check Cauchy-Riemann equations.** The Cauchy-Riemann equations state that: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \] Substituting the values: 1. \( 1 \neq -1 \) (not satisfied) 2. \( 0 \neq 0 \) (satisfied) Since the first equation is not satisfied, \( f(z) = \bar{z} \) is not differentiable at any point. ### 3. Show that \( u = x^3 - 3xy^2 \) is a harmonic function. A function \( u(x, y) \) is harmonic if it satisfies Laplace's equation: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \] **Step 1: Calculate the first derivatives.** \[ \frac{\partial u}{\partial x} = 3x^2 - 3y^2 \] \[ \frac{\partial u}{\partial y} = -6xy \] **Step 2: Calculate the second derivatives.** \[ \frac{\partial^2 u}{\partial x^2} = 6x \] \[ \frac{\partial^2 u}{\partial y^2} = -6x \] **Step 3: Check Laplace's equation.** \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 6x - 6x = 0 \] Since Laplace's equation is satisfied, \( u \) is a harmonic function. ### 4. Show that \( u = x^3 - 3xy^2 \) is an analytic function. To show that \( u \) is analytic, we need to find a corresponding \( v \) such that \( f(z) = u + iv \) is differentiable. **Step 1: Find \( v \) using the Cauchy-Riemann equations.** From the previous part, we have: \[ \frac{\partial u}{\partial x} = 3x^2 - 3y^2, \quad \frac{\partial u}{\partial y} = -6xy \] Using Cauchy-Riemann equations: 1. \( \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} = 6xy \) 2. \( \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = 3x^2 - 3y^2 \) **Step 2: Integrate to find \( v \).** Integrating \( \frac{\partial v}{\partial x} \): \[ v = 3xy^2 + g(y) \] Now differentiate with respect to \( y \): \[ \frac{\partial v}{\partial y} = 6xy + g'(y) \] Setting this equal to \( 3x^2 - 3y^2 \): \[ 6xy + g'(y) = 3x^2 - 3y^2 \] This implies \( g'(y) = -3y^2 \), so integrating gives: \[ g(y) = -y^3 + C \] Thus, \( v = 3xy^2 - y^3 + C \). ### Radius of Convergence of the Power Series To find the radius of convergence of a power series, we typically use the formula: \[ R = \frac{1}{\limsup_{n \to \infty} |a