MY-ASSIGNMENTEXPERT™可以为您提供Michigan State University MTH829 Complex Analysis复分析的代写代考和辅导服务！
这是密歇根州立大学复分析课程的代写成功案例。
MTH829课程简介
Description
Welcome to MTH 829 Complex Analysis. Please follow the link below to find the syllabus. Announcements about the course will be posted here. The on-line forum for asking and answering questions can be found at the Q&A tab above.
To use the forum you will need to register at the link you will have received in your msu email. If you have not used piazza before with your msu email, you will be asked to create a password the first time you log in.
Prerequisites
PREREQUISITES: Solid background in Real Analysis and Multiple Dimensional Analysis, ability in doing rigorous proofs,DESCRIPTION: The course will cover most of Chapters I-VIII. Here is a brief list of the main topics: complex numbers and elementary topology of the complex plane, power series, uniformly convergence, Cauchy’s theorem, winding number, Laurent series, residues, contour integrals, evaluation of definite integrals, conformal mappings, harmonic functions, and other topics if time permits.
HOMEWORK: This will be the most important part of your learning experience in the course! Homework problems will be assigned in class and collected roughly every week. Problem sets must be written up neatly and logically, with appropriate explanations provided. You can use my office hours to discuss any difficulties you may be having with these problems. Exams will consist primarily of problems similar to examples and homework problems.
MTH829 Complex Analysis HELP（EXAM HELP， ONLINE TUTOR）
(a) Note that $z=\frac{\pi}{2}+k \pi, k$ is any integer, are simple pole of $\tan z$ since
$$
\begin{aligned}
& \sin \left(\frac{\pi}{2}+k \pi\right)=(-1)^k \neq 0 . \
& \text { Now, } \operatorname{Res}\left(\tan z, \frac{\pi}{2}+k \pi\right)=\frac{\sin \left(\frac{\pi}{2}+k \pi\right)}{\left.\frac{d}{d z} \cos z\right|_{z=\frac{\pi}{2}+k \pi}}=-1 \text {. } \
&
\end{aligned}
$$
(b) Obviously, $z=1$ is a pole of order $n$. We have
$$
\begin{aligned}
\operatorname{Res}(f, 1) & =\frac{1}{(n-1) !} \frac{d^{n-1}}{d z^{n-1}}\left[(z-1)^n \frac{z^{2 n}}{(z-1)^n}\right] \
& =\frac{(2 n) !}{(n-1) !(n+1) !} .
\end{aligned}
$$
The point $z=0$ is a removable singularity of $f(z)$ since the Laurent expansion of $f(z)$ valid in the region $|z|>0$ is given by
$$
f(z)=z-\frac{z^3}{3 !}+\frac{z^5}{5 !}-\cdots, \quad|z|>0 .
$$
The function $f$ is simply defined “incorrectly” at $z=0$.
$$
\operatorname{Res}(f, 0)=\text { coefficient of } \frac{1}{z} \text { in the above Laurent series }=0 .
$$
First, consider the Taylor series expansion of $2 \cos z-2+z^2$ at $z=0$ :
$$
\begin{aligned}
2 \cos z-2+z^2 & =2\left[1-\frac{z^2}{2}+\frac{z^4}{4 !}-\frac{z^6}{6 !}+\cdots\right]-2+z^2 \
& =\frac{z^4}{12}\left(1-\frac{z^2}{30}+\cdots\right)
\end{aligned}
$$
Consider
$$
\lim {z \rightarrow 0} z^8 f=\lim {z \rightarrow 0} \frac{12^2}{1-\frac{z^2}{30}+\cdots}=144,
$$
so that $f$ has a pole of order 8 at $z=0$. Since $f$ is even so that $\operatorname{Res}(f, 0)=-\operatorname{Res}(f, 0)$; hence,
$$
\operatorname{Res}(f, 0)=0
$$
(a) Note that $z=1$ is a double pole of the integrand and it is the only pole included inside $|z|=2$. We then have
$$
\begin{aligned}
\oint_{|z|=2} \frac{z^4+z}{(z-1)^2} d z & =2 \pi i \operatorname{Res}\left(\frac{z^4+z}{(z-1)^2}, 1\right) \
& =2 \pi i \lim {z \rightarrow 1}\left(z^4+z\right)^{\prime}=2 \pi i \lim {z \rightarrow 1}\left(4 z^3+1\right)=10 \pi i .
\end{aligned}
$$
(b) Note that $z=0$ is a double pole of the integrand and it is the only pole included inside $|z|=2$. We then have
$$
\begin{aligned}
\oint_{|z|=2} \frac{z^3+3 z+1}{z^4-5 z^2} d z & =2 \pi i \operatorname{Res}\left(\frac{z^3+3 z+1}{z^4-5 z^2}, 0\right) \
& =2 \pi i \lim {z \rightarrow 0}\left(\frac{z^3+3 z+1}{z^2-5}\right)^{\prime} \ & =2 \pi i \lim {z \rightarrow 0} \frac{\left(z^2-5\right)\left(3 z^2+3\right)-\left(z^2+3 z+1\right)(2 z)}{\left(z^2-5\right)^2}=\frac{-6 \pi i}{5}
\end{aligned}
$$
(c) Note that $z=0$ is a double pole of $\sinh ^2 z / z^4$. Hence
$$
\begin{aligned}
\oint_{|z|=2} \frac{\sinh ^2 z}{z^4} d z & =2 \pi i \operatorname{Res}\left(\frac{\sinh ^2 z}{z^4}, 0\right) \
& =2 \pi i \lim {z \rightarrow 0} \frac{d}{d z}\left[\frac{\sinh ^2 z}{z^2}\right] \ & =2 \pi i \lim {z \rightarrow 0} \frac{d}{d z}\left[\frac{\left(z+\frac{z^3}{3 !}+\cdots\right)^2}{z^2}\right]=0 .
\end{aligned}
$$
(d) Consider
$$
\begin{aligned}
\oint_{|z-i|=2} \frac{e^z+z}{(z-1)^4} d z & =2 \pi i \operatorname{Res}\left(\frac{e^z+z}{(z-1)^4}, 1\right) \
& =2 \pi i \lim _{z \rightarrow 1} \frac{\left(e^z+z\right)^{\prime \prime \prime}}{3 !}=2 \pi i\left(\frac{e}{6}\right)=\frac{\pi e i}{3}
\end{aligned}
$$
MY-ASSIGNMENTEXPERT™可以为您提供MICHIGAN STATE UNIVERSITY MATH113 COMPLEX ANALYSIS复分析的代写代考和辅导服务！
