数学代写|MATH3364 complex analysis

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MATH 3364课程简介

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Prerequisites: MATH 3331.

Course Description: The complex number system, analytic functions, the Cauchy integral theorem, series representation, residue theory, and conformal mapping

Text: Fundamentals of Complex Analysis with Applications to Engineering and Science, 3rd Edition, by E.B. Saff and A.D. Snider, Prentice-Hall, 2003. ISBN: 9780139078743

Prerequisites 

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MATH3364 COMPLEX ANALYSIS HELP（EXAM HELP， ONLINE TUTOR）

Consider the principal branch of the complex logarithm $\log z$, defined on the complex plane with a branch cut along the non-positive real axis. Then, for any $z$ in the domain, we can write $z = re^{i\theta}$, where $r > 0$ and $-\pi < \theta < \pi$. We define $f(z) = e^{z\log z}$.

Since $z = x$ on the positive real axis, we have $f(x) = e^{x\log x} = x^x$ for $x > 0$. Thus, $f$ agrees with the given function on the positive real axis.

To find $f(i)$, we have $i = e^{i\pi/2}$, so $f(i) = e^{i\pi/2 \cdot \log i}$. Using the properties of the principal branch of the logarithm, we have $\log i = \ln|i| + i\arg i = i\pi/2$, so $f(i) = e^{i\pi/2 \cdot i\pi/2} = e^{-\pi/4}(\cos(\pi/4) + i\sin(\pi/4)) = e^{-\pi/4}(1 + i)$.

Similarly, we have $-i = e^{-i\pi/2}$, so $f(-i) = e^{-i\pi/2 \cdot \log(-i)}$. Again using the properties of the principal branch of the logarithm, we have $\log(-i) = \ln|-i| + i\arg(-i) = i\pi/2$, so $f(-i) = e^{-i\pi/2 \cdot i\pi/2} = e^{\pi/4}(\cos(\pi/4) – i\sin(\pi/4)) = e^{\pi/4}(1 – i)$.

To show that $f(\bar{z}) = \overline{f(z)}$ for all $z$ in the domain, we have $\bar{z} = re^{-i\theta}$, so $f(\bar{z}) = e^{\bar{z}\log \bar{z}} = e^{re^{-i\theta}(\ln|r| + i(-\theta+\pi))} = e^{r\ln|r| – ir\theta}(\cos(r(\pi-\theta)) + i\sin(r(\pi-\theta)))$. On the other hand, $\overline{f(z)} = \overline{e^{z\log z}} = e^{\bar{z}\overline{\log z}} = e^{re^{-i\theta}(\ln|r| – i\theta)} = e^{r\ln|r| + ir\theta}(\cos(r\theta) – i\sin(r\theta))$. It is clear that the real parts of $f(\bar{z})$ and $\overline{f(z)}$ are equal, so we just need to show that the imaginary parts are equal. We have $\sin(r(\pi-\theta)) = \sin(r\theta)$ and $\cos(r(\pi-\theta)) = -\cos(r\theta)$, so the imaginary parts of $f(\bar{z})$ and $\overline{f(z)}$ are equal. Therefore, $f(\bar{z}) = \overline{f(z)}$ for all $z$ in the domain, as desired.

To prove that $f(z)=\frac{\cos z-1}{z^2}$ has a removable singularity at $z=0$, we need to show that $f(z)$ can be defined at $z=0$ in such a way that it is analytic (holomorphic) in a neighborhood of $z=0$.

We first note that $\cos z – 1 = -\frac{z^2}{2} + O(z^4)$, where $O(z^4)$ denotes terms of degree at least 4 in $z$. Thus, we can write

$$ f(z) = \frac{\cos z – 1}{z^2} = -\frac{1}{2} + \frac{O(z^2)}{z^2}. $$

As $z$ approaches $0$, the second term goes to zero faster than the first term, so $f(z)$ has a finite limit as $z$ approaches $0$, namely $-\frac{1}{2}$. Thus, we can define $f(0)=-\frac{1}{2}$ to extend $f(z)$ to a continuous function at $z=0$.

Now we need to show that $f(z)$ is analytic in a neighborhood of $z=0$. Let $D_r(0)$ denote the disk of radius $r$ centered at $z=0$. We want to show that $f(z)$ has a power series expansion around $z=0$, which is the definition of analyticity.

For $z \neq 0$, we can write

$$ \cos z – 1 = -\frac{z^2}{2} + \sum_{n=2}^\infty \frac{(-1)^n z^{2n}}{(2n)!}, $$

so

$$ f(z) = \frac{\cos z – 1}{z^2} = -\frac{1}{2} + \sum_{n=2}^\infty \frac{(-1)^n z^{2(n-1)}}{(2n)!}. $$

This is a power series expansion of $f(z)$ around $z=0$, so $f(z)$ is analytic in $D_r(0)$ for any $r>0$. Therefore, $f(z)$ has a removable singularity at $z=0$.

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