Cauchy's Theorem and Cauchy's Formula Exercises

[Tsunoyukami]

I am studying for a test I have tomorrow evening and have finally reached the final section. I'm not 100% whether or not I am approaching these questions correctly or not so please bear with me. I have four questions and am not sure whether it merits starting separate threads considering they are similar - therefore, I will attempt all here and will only start separate threads if requested. Thanks a bunch in advance for any and all assistance!

Homework Statement

"In Exercises 1 to 4, evaluate the given integral using Cauchy's Formula or Theorem.

1. $\int_{|z| = 1} \frac{z}{(z-2)^{2}} dz$
2. $\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz$
3. $\int_{|z+1| = 2} \frac{z^{2}}{4-z^{2}} dz$
4. $\int_{|z|=1} \frac{sin(z)}{z} dz$" (Complex Variables, 2nd Edition by Stephen D. Fisher; pg. 116)

Homework Equations

"Cauchy's Theorem Suppose that f is analytic on a domain D. Let $\gamma$ be a piecewise smooth simple closed curve in D whose inside Ωalso lies in D. Then

$$\int_{\gamma} f(z) dz = 0$$" (Complex Variables, 2nd Edition by Stephen D. Fisher; pg. 106)


"Cauchy's Formula Suppose that f is analytic on a domain D and that $\gamma$ is a piecewise smooth, positively oriented simple closed curve in D whose inside Ω also lies in D. Then

$$f(z) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(a)}{(a - z)} da$$ for all $z \in$ Ω" (Complex Variables, 2nd Edition by Stephen D. Fisher; pg. 111)

The Attempt at a Solution

Exercise 1:

$$\int_{|z| = 1} \frac{z}{(z-2)^{2}} dz$$

For this exercise, $\gamma = |z| = 1$ and so this integral cannot be rewritten in the form required by Cauchy's Formula. We use Cauchy's Theorem instead. Therefore,

$$\int_{|z| = 1} \frac{z}{(z-2)^{2}} dz = 0$$


Exercise 2:

Here $\gamma = |z| = 2$ and so we can write this integral in the form of Cauchy's Formula:

$$\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz$$
$$\int_{|z| = 2} \frac{e^{z}}{(z-0)(z-3)} dz$$
$$\int_{|z| = 2} \frac{g(z)}{(z-0)} dz, g(z) = \frac{e^{z}}{z-3}$$
$$\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz = (2\pi i)g(0) = (2\pi i)\frac{e^{0}}{0-3} = (2\pi i)\frac{-1}{3}$$

$$\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz = - \frac{2\pi i}{3} $$

Exercise 3:

This exercise has $\gamma = |z+1| = 2$.

$$\int_{|z+1| = 2} \frac{z^{2}}{4-z^{2}} dz$$
$$\int_{|z+1| = 2} \frac{z^{2}}{(2-z)(2+z)} dz$$
$$\int_{|z+1| = 2} \frac{z^{2}}{-(z-2)(z+2)} dz$$

We find $g(z) = \frac{z^{2}}{-(z-2)} = \frac{z^{2}}{2-z}$ and evaluate $g(z)$ at $z = -2)$:

$$\int_{|z+1| = 2} \frac{z^{2}}{4-z^{2}} dz = (2\pi i)g(-2) = (2\pi i)\frac{(-2)^{2}}{2-(-2)} = (2\pi i)\frac{4}{4} = 2\pi i$$


Exercise 4:

Here we have $\gamma = |z| = 1$ and:

$$\int_{|z|=1} \frac{sin(z)}{z} dz$$
$$\int_{|z|=1} \frac{sin(z)}{z-0} dz$$

We say $g(z) = sin(z)$ so we can write:

$$\int_{|z|=1} \frac{sin(z)}{z-0} dz = (2\pi i)g(0) = (2\pi i)sin(0) = 0$$


Is this the proper approach to solving these four problems? I get the same final answer for Exercises 1 and 3, but the answers to 2 and 4 are not provided. Thanks very much for your time! :)

 

[Dick]

I am studying for a test I have tomorrow evening and have finally reached the final section. I'm not 100% whether or not I am approaching these questions correctly or not so please bear with me. I have four questions and am not sure whether it merits starting separate threads considering they are similar - therefore, I will attempt all here and will only start separate threads if requested. Thanks a bunch in advance for any and all assistance!

Homework Statement

"In Exercises 1 to 4, evaluate the given integral using Cauchy's Formula or Theorem.

1. $\int_{|z| = 1} \frac{z}{(z-2)^{2}} dz$
2. $\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz$
3. $\int_{|z+1| = 2} \frac{z^{2}}{4-z^{2}} dz$
4. $\int_{|z|=1} \frac{sin(z)}{z} dz$" (Complex Variables, 2nd Edition by Stephen D. Fisher; pg. 116)

Homework Equations

"Cauchy's Theorem Suppose that f is analytic on a domain D. Let $\gamma$ be a piecewise smooth simple closed curve in D whose inside Ωalso lies in D. Then

$$\int_{\gamma} f(z) dz = 0$$" (Complex Variables, 2nd Edition by Stephen D. Fisher; pg. 106)


"Cauchy's Formula Suppose that f is analytic on a domain D and that $\gamma$ is a piecewise smooth, positively oriented simple closed curve in D whose inside Ω also lies in D. Then

$$f(z) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(a)}{(a - z)} da$$ for all $z \in$ Ω" (Complex Variables, 2nd Edition by Stephen D. Fisher; pg. 111)

The Attempt at a Solution

Exercise 1:

$$\int_{|z| = 1} \frac{z}{(z-2)^{2}} dz$$

For this exercise, $\gamma = |z| = 1$ and so this integral cannot be rewritten in the form required by Cauchy's Formula. We use Cauchy's Theorem instead. Therefore,

$$\int_{|z| = 1} \frac{z}{(z-2)^{2}} dz = 0$$


Exercise 2:

Here $\gamma = |z| = 2$ and so we can write this integral in the form of Cauchy's Formula:

$$\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz$$
$$\int_{|z| = 2} \frac{e^{z}}{(z-0)(z-3)} dz$$
$$\int_{|z| = 2} \frac{g(z)}{(z-0)} dz, g(z) = \frac{e^{z}}{z-3}$$
$$\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz = (2\pi i)g(0) = (2\pi i)\frac{e^{0}}{0-3} = (2\pi i)\frac{-1}{3}$$

$$\int_{|z| = 2} \frac{e^{z}}{z(z-3)} dz = - \frac{2\pi i}{3} $$

Exercise 3:

This exercise has $\gamma = |z+1| = 2$.

$$\int_{|z+1| = 2} \frac{z^{2}}{4-z^{2}} dz$$
$$\int_{|z+1| = 2} \frac{z^{2}}{(2-z)(2+z)} dz$$
$$\int_{|z+1| = 2} \frac{z^{2}}{-(z-2)(z+2)} dz$$

We find $g(z) = \frac{z^{2}}{-(z-2)} = \frac{z^{2}}{2-z}$ and evaluate $g(z)$ at $z = -2)$:

$$\int_{|z+1| = 2} \frac{z^{2}}{4-z^{2}} dz = (2\pi i)g(-2) = (2\pi i)\frac{(-2)^{2}}{2-(-2)} = (2\pi i)\frac{4}{4} = 2\pi i$$


Exercise 4:

Here we have $\gamma = |z| = 1$ and:

$$\int_{|z|=1} \frac{sin(z)}{z} dz$$
$$\int_{|z|=1} \frac{sin(z)}{z-0} dz$$

We say $g(z) = sin(z)$ so we can write:

$$\int_{|z|=1} \frac{sin(z)}{z-0} dz = (2\pi i)g(0) = (2\pi i)sin(0) = 0$$


Is this the proper approach to solving these four problems? I get the same final answer for Exercises 1 and 3, but the answers to 2 and 4 are not provided. Thanks very much for your time! :)

I only checked 2 and 4 and skimmed over 1 and 3, but you seem to have a pretty good idea of what you are doing and those are correct. Well done!

 

[Tsunoyukami]

Thank you very much for the prompt reply. It's difficult to be sure I understand when there are only four exercises (directly) related to this material in the book.