Tsunoyukami
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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!
"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)
"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)
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! :)
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! :)