# 代数|MA20217/MA20219 Algebra 2B代写

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Elementary axiomatic theory of rings. Integral domains, fields, characteristic. Subrings and product of rings. Homomorphisms, ideals and quotient rings. Isomorphism theorems. Fields of fractions.

Use Cramer’s rule to solve the system
\begin{aligned} 2 x_{1}-x_{2} &=1 \ 4 x_{1}+4 x_{2} &=20 . \end{aligned}
Solution. The coefficient matrix and right-hand-side vectors are
$$A=\left[\begin{array}{rr} 2 & -1 \ 4 & 4 \end{array}\right] \text { and } \mathbf{b}=\left[\begin{array}{r} 1 \ 20 \end{array}\right]$$
so that
$$\operatorname{det} A=8-(-4)=12$$
and therefore
$$x_{1}=\frac{\left|\begin{array}{rr} 2 & 1 \ 4 & 20 \end{array}\right|}{\left|\begin{array}{rr} 2 & -1 \ 4 & 4 \end{array}\right|}=\frac{36}{12}=3 \quad \text { and } \quad x_{2}=\frac{\left|\begin{array}{rr} 1 & -1 \ 20 & 4 \end{array}\right|}{\left|\begin{array}{rr} 2 & -1 \ 4 & 4 \end{array}\right|}=\frac{24}{12}=2$$

## MA20217/MA20219 COURSE NOTES ：

Since this holds for all $x$, we conclude that $(f+g)+h=f+(g+h)$, which is the associative law for addition of vectors.

Next, if 0 denotes the constant function with value 0 , then for any $f \in V$ we have that for all $0 \leq x \leq 1$,
$$(f+0)(x)=f(x)+0=f(x) .$$
(We don’t write the zero element of this vector space in boldface because it’s customary not to write functions in bold.) Since this is true for all $x$ we have that $f+0=f$, which establishes the additive identity law. Also, we define $(-f)(x)=-(f(x))$ so that for all $0 \leq x \leq 1$,
$$(f+(-f))(x)=f(x)-f(x)=0,$$

# 数学分析|MA20219 Analysis 2B代写

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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).

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 $z \in B_{2}(0) \backslash{0}$
$$\frac{1}{z(z-2)}=\sum_{k=0}^{\infty} \frac{-1}{2^{k+1}} z^{k-1}=\sum_{n=-1}^{\infty} \frac{-1}{2^{n+2}} z^{n}$$