If we adapt the ideas seen in previous analysis units, and in particular the idea of a derivative, to complex variables, then we find a remarkably beautiful theory. In particular, a lot of the complications coming from limited differentiability or from various notions of convergence fall away. Of course, this comes at a price, which is that complex differentiability is much more restrictive than the real counterpart. So we will have a very powerful theory, but it will apply only to a relatively small set of functions (which still includes rational, exponential, and trigonometric functions and much more).

这是一份Bath巴斯大学学院MA20219作业代写的成功案

Let $U=\mathbb{C} \backslash{1-i,-1-i, \log 2+\pi i}$. Consider the function $f: U \rightarrow \mathbb{C}$ given by
$$
f(z)=\frac{e^{z}+2}{(z-\log 2-\pi i)\left(z^{2}+2 i z-2\right)}, \quad z \in U .
$$
Let $\gamma:[-\pi / 2, \pi / 2+20] \rightarrow \mathbb{C}$ such that
$$
\gamma(t)= \begin{cases}10 e^{i t} & \text { if }-\pi / 2 \leq t \leq \pi / 2 \ (10+\pi / 2-t) i & \text { if } \pi / 2<t \leq \pi / 2+20\end{cases}
$$
Set $\Gamma=\gamma([-\pi / 2, \pi / 2+20])$.
You may use any result from the lectures and exercises to answer the following questions. If you need to determine any winding numbers, you may use geometric intuition.
(a) Show that $f$ has poles of order 1 at $1-i$ and $-1-i$.
(b) Sketch the curve $\Gamma$, including its position relative to the points $1-i$ and $-1-i$. Indicate the orientation of $\gamma$.
(c) The singularity of $f$ at $\log 2+\pi i$ is removable. Using this information (for which no proof is required), or otherwise, find
$$
\int_{\gamma} f(z) d z .
$$

MA20219 COURSE NOTES ：

i. The singularities of the given function are 0 and 2 . The distance from the centre to the nearest singularity (apart from the centre itself) is therefore 2. So by a result from the lecture notes, the punctured disc of convergence has radius 2 and therefore is $B_{2}(0) \backslash{0}$.
ii. We have (using the geometric series) for $z \in B_{2}(0)$
$$
\frac{1}{z-2}=\frac{-1}{2} \frac{1}{1-\frac{z}{2}}=\frac{-1}{2} \sum_{k=0}^{\infty}\left(\frac{z}{2}\right)^{k}=\sum_{k=0}^{\infty} \frac{-1}{2^{k+1}} z^{k}
$$
so that for $z \in B_{2}(0) \backslash{0}$
$$
\frac{1}{z(z-2)}=\sum_{k=0}^{\infty} \frac{-1}{2^{k+1}} z^{k-1}=\sum_{n=-1}^{\infty} \frac{-1}{2^{n+2}} z^{n}
$$