Find the value of मान ज्ञात कीजिए : (a) \( \int_{|:|=1} \frac{e^{2 z}}{(z+1)^{4}} d z \). (b) \( \int_{|:|=1} \frac{\sin ^{6} z}{\left(z-\frac{\pi}{6}\right)^{3}} d z \). Discuss the singularities of the following functions (a) \( f(z)=\frac{1}{z\left(e^{2}-1\right)} \). (b) \( f(z)=\tan \frac{1}{z} \). (c) \( f(z)=\frac{z}{(z-1)^{2}} \cos \frac{1}{z-2} \). fिम्नलिखित फलनों की विवित्रताओं की विवेचना कीजिए : (a) \( f(z)=\frac{1}{z\left(e^{2}-1\right)} \). (b) \( f(z)=\tan \frac{1}{z} \). (c) \( f(z)=\frac{z}{(z-1)^{2}} \cos \frac{1}{z-2} \).

To solve the integrals and discuss the singularities of the given functions, we will proceed step by step. ### Part (a): Evaluate the integral \( \int_{|z|=1} \frac{e^{2 z}}{(z+1)^{4}} d z \) 1. **Identify the singularities**: The integrand \( \frac{e^{2 z}}{(z+1)^{4}} \) has a singularity at \( z = -1 \). This is a pole of order 4. 2. **Residue Calculation**: To evaluate the integral, we can use the residue theorem. We need to find the residue of the function at the pole \( z = -1 \). The residue of a function \( f(z) \) at a pole of order \( n \) can be calculated using the formula: \[ \text{Residue} = \frac{1}{(n-1)!} \lim_{z \to z_0} \frac{d^{n-1}}{dz^{n-1}} \left( (z - z_0)^{n} f(z) \right) \] Here, \( n = 4 \) and \( z_0 = -1 \). We need to compute: \[ \text{Residue} = \frac{1}{3!} \lim_{z \to -1} \frac{d^{3}}{dz^{3}} \left( (z + 1)^{4} \frac{e^{2 z}}{(z + 1)^{4}} \right) = \frac{1}{6} \lim_{z \to -1} \frac{d^{3}}{dz^{3}} e^{2 z} \] 3. **Calculate the derivatives**: \[ e^{2 z} \text{ and its derivatives: } \] - First derivative: \( \frac{d}{dz} e^{2 z} = 2 e^{2 z} \) - Second derivative: \( \frac{d^{2}}{dz^{2}} e^{2 z} = 4 e^{2 z} \) - Third derivative: \( \frac{d^{3}}{dz^{3}} e^{2 z} = 8 e^{2 z} \) Evaluating at \( z = -1 \): \[ \frac{d^{3}}{dz^{3}} e^{2 z} \bigg|_{z=-1} = 8 e^{-2} \] 4. **Final residue**: \[ \text{Residue} = \frac{1}{6} \cdot 8 e^{-2} = \frac{4 e^{-2}}{3} \] 5. **Integral evaluation**: By the residue theorem: \[ \int_{|z|=1} \frac{e^{2 z}}{(z+1)^{4}} d z = 2 \pi i \cdot \text{Residue} = 2 \pi i \cdot \frac{4 e^{-2}}{3} = \frac{8 \pi i e^{-2}}{3} \] ### Part (b): Evaluate the integral \( \int_{|z|=1} \frac{\sin^{6} z}{\left(z - \frac{\pi}{6}\right)^{3}} d z \) 1. **Identify the singularities**: The integrand \( \frac{\sin^{6} z}{\left(z - \frac{\pi}{6}\right)^{3}} \) has a singularity at \( z = \frac{\pi}{6} \). This is a pole of order 3. 2. **Residue Calculation**: We need to find the residue at \( z = \frac{\pi}{6} \). Using the same residue formula: \[ \text{Residue} = \frac{1}{2!} \lim_{z \to \frac{\pi}{6}} \frac{d^{2}}{dz^{2}} \left( (z - \frac{\pi}{6})^{3} \frac{\sin^{6} z}{(z - \frac{\pi}{6})^{3}} \right) = \frac{1}{2} \lim_{z \to \frac{\pi}{6}} \frac{d^{2}}{dz^{2}} \sin^{6} z \] 3. **Calculate the derivatives**: - First derivative: \( \frac{d}{dz} \sin^{6} z = 6 \sin^{5} z \cos z \) - Second derivative: Using the product rule: \[ \frac{d^{2}}{dz^{2}} \sin^{6} z = 6 \left( 5 \sin^{4} z \cos^{2} z - \sin^{6} z \right) \] Evaluating at \( z = \frac{\pi}{6} \): \[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \] \[ \sin^{4}\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^{4} = \frac{1}{16}, \quad \cos^{2}\left(\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3}{4} \] \[ \frac{d^{2}}{dz^{2}} \sin^{6} z \bigg|_{z=\frac{\pi}{6}} = 6 \left( 5 \cdot \frac{1}{16} \cdot \frac{3}{4} - \left(\frac{1}{2}\right)^{6} \right) = 6 \left( \frac{15}{64} - \frac{1}{64} \right) = 6 \cdot \frac{14}{64} = \frac{21}{16} \] 4. **Final residue**: \[ \text{Residue} = \frac{1}{2} \cdot \frac{21}{16} = \frac{21}{32} \] 5. **Integral evaluation**: \[ \int_{|z|=1} \frac{\sin^{6} z}{\left(z - \frac{\pi}{6}\right)^{3}} d z = 2 \pi i \cdot \text{Residue} = 2 \pi i \cdot \frac{21}{32}