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

Question

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

UpStudy AI · Accepted Answer

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}
The integrals are evaluated as follows: (a) $$ \int_{|z|=1} \frac{e^{2 z}}{(z+1)^{4}} d z = \frac{8 \pi i e^{-2}}{3} $$ (b) $$ \int_{|z|=1} \frac{\sin^{6} z}{\left(z - \frac{\pi}{6}\right)^{3}} d z = \frac{21 \pi i}{16} $$ ### Singularities Discussion (a) $$ f(z) = \frac{1}{z(e^{2} - 1)} $$ - **Singularities**: The function has a pole at $$ z = 0 $$ and a pole at $$ z = \frac{2\pi i}{2} = \pi i $$ (since $$ e^{2} = 1 $$ when $$ z = \pi i $$). - **Residues**: - At $$ z = 0 $$: $$ \text{Residue} = \lim_{z \to 0} z \cdot \frac{1}{z(e^{2} - 1)} = \frac{1}{e^{2} - 1} $$ - At $$ z = \pi i $$: $$ \text{Residue} = \lim_{z \to \pi i} (z - \pi i) \cdot \frac{1}{z(e^{2} - 1)} = \frac{1}{\pi i (e^{2} - 1)} $$ (b) $$ f(z) = \tan \frac{1}{z} $$ - **Singularities**: The function has an essential singularity at $$ z = 0 $$ because $$ \frac{1}{z} $$ approaches infinity as $$ z $$ approaches 0. - **Residues**: The residue at $$ z = 0 $$ is more complex and involves the Laurent series expansion of $$ \tan \frac{1}{z} $$, which has infinitely many terms. (c) $$ f(z) = \frac{z}{(z-1)^{2}} \cos \frac{1}{z-2} $$ - **Singularities**: The function has a pole of order 2 at $$ z = 1 $$ and an essential singularity at $$ z = 2 $$ due to $$ \cos \frac{1}{z-2} $$. - **Residues**: - At $$ z = 1 $$: $$ \text{Residue} = \lim_{z \to 1} (z - 1)^2 \cdot \frac{z}{(z-1)^{2}} \cos \frac{1}{z-2} = \cos \frac{1}{-1} = \cos(-1) = \cos(1) $$ - At $$ z = 2 $$: $$ \text{Residue} $$ is determined by the Laurent series expansion of $$ \cos \frac{1}{z-2} $$, which has infinitely many terms. **Summary**: - **Integral (a)**: $$ \frac{8 \pi i e^{-2}}{3} $$ - **Integral (b)**: $$ \frac{21 \pi i}{16} $$ - **Singularities**: - $$ f(z) = \frac{1}{z(e^{2} - 1)} $$: Poles at $$ z = 0 $$ and $$ z = \pi i $$ - $$ f(z) = \tan \frac{1}{z} $$: Essential singularity at $$ z = 0 $$ - $$ f(z) = \frac{z}{(z-1)^{2}} \cos \frac{1}{z-2} $$: Pole of order 2 at $$ z = 1 $$ and essential singularity at $$ z = 2 $$

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