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MAT334--Misc / Re: Test2
« on: December 05, 2018, 01:15:02 PM »
Actually there could be no comments.
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MAT334--Misc / Re: bonus marks from lectures and tutorials?
« on: December 02, 2018, 08:12:43 PM »4
End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 5A
« on: November 27, 2018, 11:19:47 AM »And how do you prove that two last equations are incompatible?I don't know if I fully understood what you mean but here's my attempt:
Suppose $$f'(z_0)=10z^4+4=0 \\ \Rightarrow z_0^4=-\frac{2}{5} \Rightarrow z_0=\frac{2}{5}e^{i\frac{\pi + 2k\pi}{4}}$$
$$\begin{align}f(z_0)&=2z_0\cdot z_0^4+4z_0+1=0\\&=2z_0\cdot(-\frac{2}{5})+4z_0+1=0\\&=\frac{16}{5}z_0+1=0\\ \Rightarrow z_0=-\frac{16}{5}\neq \frac{2}{5}e^{i\frac{\pi + 2k\pi}{4}}\end{align}$$
5
End of Semester Bonus--sample problem for FE / Re: FE Sample--Problem 5A
« on: November 27, 2018, 10:36:20 AM »
$(a)$ $\\$
At $|z|=1$, $$\begin{align}|2z^5+4z+1+(-4z)|&=|2z^5+1|\\&\leq |2z^5|+1\\&=3\\&<4=|-4z|\end{align}$$
By Rouche's Theorem,
$2z^5+4z+1$ and $-4z$ has the same number of zeros.$\\$ Since $-4z$ has $1$ zero, therefore $2z^5+4z+1$ has $1$ zero in the disk $\{z\colon |z|<1\}$.
$\\$
$\\$
$(b)$ $\\$
At $|z|=2$, $$\begin{align}|2z^5+4z+1+(-2z^5)|&=|4z+1|\\&\leq |4z|+1\\&=5\\&<64=|-2z^5|\end{align}$$
By Rouche's Theorem, $2z^5+4z+1$ and $-2z^5$ has the same number of zeros.$\\$
Since $-2z^5$ has $5$ zeros inside $|z|=2$, therefore $2z^5+4z+1$ has $4$ zeros $(5-1=4)$ in the annulus $\{z\colon 1 <|z| < 2\}$.
$\\$
$\\$
$(c)$ $\\$
Notice that the degree of $f(z)=2z^5+4z+1$ is $5$. Which means it has at most $5$ roots. Now from part(b), all the roots are inside $|z|=2$, therefore there are no roots/zeros in the domain $\{z\colon |z|>2\}$.
$\\$
$\\$
$\\$
Show distinct:
$$f(z_0)=2z_{0}^5+4z_0+1=0 \\ f'(z_0)=10z_0^4+4\neq0$$
Thus the multiplicity is $1$, therefore they are all distinct.
At $|z|=1$, $$\begin{align}|2z^5+4z+1+(-4z)|&=|2z^5+1|\\&\leq |2z^5|+1\\&=3\\&<4=|-4z|\end{align}$$
By Rouche's Theorem,
$2z^5+4z+1$ and $-4z$ has the same number of zeros.$\\$ Since $-4z$ has $1$ zero, therefore $2z^5+4z+1$ has $1$ zero in the disk $\{z\colon |z|<1\}$.
$\\$
$\\$
$(b)$ $\\$
At $|z|=2$, $$\begin{align}|2z^5+4z+1+(-2z^5)|&=|4z+1|\\&\leq |4z|+1\\&=5\\&<64=|-2z^5|\end{align}$$
By Rouche's Theorem, $2z^5+4z+1$ and $-2z^5$ has the same number of zeros.$\\$
Since $-2z^5$ has $5$ zeros inside $|z|=2$, therefore $2z^5+4z+1$ has $4$ zeros $(5-1=4)$ in the annulus $\{z\colon 1 <|z| < 2\}$.
$\\$
$\\$
$(c)$ $\\$
Notice that the degree of $f(z)=2z^5+4z+1$ is $5$. Which means it has at most $5$ roots. Now from part(b), all the roots are inside $|z|=2$, therefore there are no roots/zeros in the domain $\{z\colon |z|>2\}$.
$\\$
$\\$
$\\$
Show distinct:
$$f(z_0)=2z_{0}^5+4z_0+1=0 \\ f'(z_0)=10z_0^4+4\neq0$$
Thus the multiplicity is $1$, therefore they are all distinct.
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MAT334--Misc / How to type properly: MathJax basic tutorial and quick reference
« on: November 21, 2018, 09:19:58 PM »
MathJax basic tutorial and quick reference
Personally, I prefer to use https://www.mathjax.org/
It is very much straightforward. Scroll down and click the green button says "Try a live demo" then you can begin typing your solution out and your typing preview will be shown right below it.
Below is a link that contains all the helpful tips/references which I found is pretty much enough for typing all kinds solutions out perfectly.
https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
You can use Crtl+F (Windows keyboard) or Command + F (Mac keyboard) to search.
If there are any problems\questions relating to typing your solutions, feel free to send PMs.
Personally, I prefer to use https://www.mathjax.org/
It is very much straightforward. Scroll down and click the green button says "Try a live demo" then you can begin typing your solution out and your typing preview will be shown right below it.
Below is a link that contains all the helpful tips/references which I found is pretty much enough for typing all kinds solutions out perfectly.
https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
You can use Crtl+F (Windows keyboard) or Command + F (Mac keyboard) to search.
If there are any problems\questions relating to typing your solutions, feel free to send PMs.
7
MAT334--Misc / Scrap paper
« on: November 20, 2018, 12:07:05 PM »
Will scrap papers be provided for test by any chance?
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Quiz-6 / Re: Q6 TUT 0301
« on: November 17, 2018, 04:12:24 PM »
$$\because\sin(z)=\sin(z-\pi+\pi)=-\sin(z-\pi), \space\space\space\space\space\space \space\space\space\space\text{since } \sin(\pi+\theta)=-\sin(\theta).$$
$$\therefore -\sin(z-\pi)=-\sum_{n=0}^\infty \frac{(-1)^n(z-\pi)^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n+1}}{(2n+1)!}$$
$$\therefore \begin{align} \frac{\sin(z)}{(z-\pi)^2}= \frac{\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n+1}}{(2n+1)!}}{(z-\pi)^2}=\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n-1}}{(2n+1)!}\end{align}$$
$$\\$$
$$\\$$
$$\frac{\sin(z)}{(z-\pi)^2}=\frac{(-1)^{n+1}(z-\pi)^{2n-1}}{(2n+1)!}=-\frac{1}{z-\pi}+\cdots$$
Therefore, the residue of $\frac{\sin(z)}{(z-\pi)^2}$ at $z_0=\pi$ is $-1$, which is the coefficient of $(z-z_0)^{-1}$.
$$\therefore -\sin(z-\pi)=-\sum_{n=0}^\infty \frac{(-1)^n(z-\pi)^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n+1}}{(2n+1)!}$$
$$\therefore \begin{align} \frac{\sin(z)}{(z-\pi)^2}= \frac{\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n+1}}{(2n+1)!}}{(z-\pi)^2}=\sum_{n=0}^\infty \frac{(-1)^{n+1}(z-\pi)^{2n-1}}{(2n+1)!}\end{align}$$
$$\\$$
$$\\$$
$$\frac{\sin(z)}{(z-\pi)^2}=\frac{(-1)^{n+1}(z-\pi)^{2n-1}}{(2n+1)!}=-\frac{1}{z-\pi}+\cdots$$
Therefore, the residue of $\frac{\sin(z)}{(z-\pi)^2}$ at $z_0=\pi$ is $-1$, which is the coefficient of $(z-z_0)^{-1}$.
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Quiz-6 / Re: Q6 TUT 0202
« on: November 17, 2018, 04:09:29 PM »
$$\because e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots$$
$$\begin{align}\therefore e^z-1&=(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)-1\\&=z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots \end{align}$$
Added omitted calculation of order of pole: (thanks Chunjing Zhang for pointing it out)
$$g(z)=\frac{1}{f(z)}=\frac{1}{\frac{1}{e^z-1}}=e^z-1\\g(z_0=0)=e^0-1=0\\g'(z)=e^z \Rightarrow g(z_0=0)=e^0=1\neq 0 \\$$
Thus the order of the pole of $f(z)$ at $z_0=0$ is $1$.
Hence we let $$\frac{1}{e^z-1}=a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots$$
$$\therefore(e^z-1)(\frac{1}{e^z-1})=1\\\therefore (z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)(a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots)=1$$
$$\Rightarrow a_{-1}+a_0z+a_1z^2+a_2z^3+\frac{1}{2}z+\frac{a_0}{2}z^2+\frac{a_1}{2}z^3+\frac{a_{-1}}{6}z^2+\frac{a_0}{6}z^3+\frac{a_{-1}}{24}z^3+\cdots=1$$
$$\therefore \begin{cases}a_{-1}=1\\a_0z+\frac{a_{-1}}{2}z=0 \Rightarrow a_0+\frac{a_{-1}}{2}=0 \Rightarrow a_0=-\frac{1}{2}\\\frac{a_{-1}}{6}z^2+\frac{a_0}{2}z^2+a_1z^2=0 \Rightarrow \frac{a_{-1}}{6}+\frac{a_0}{2}+a_1=0 \Rightarrow a_1=\frac{1}{12}\\ \frac{a_{-1}}{24}z^3+\frac{a_0}{6}z^3+\frac{a_1}{2}z^3+a_2z^3 \Rightarrow \frac{a_{-1}}{24}+\frac{a_0}{6}+\frac{a_1}{2}+a_2=0 \Rightarrow a_2=0\end{cases}$$
Therefore, the first four terms of the Laurent series:
$$\require{cancel} \cancel{1+\frac{1}{z}+(-\frac{1}{2})z^2+(0)z^3} \\ \text{Typo correction: } \frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2$$
$$\\$$
$$\\$$
The residue of given function at $z_0$ is the coefficient of $(z-z_0)^{-1}$, which is $1$.
$$\\$$
$$\\$$
If $0(z^3)=0$ is not counted since its zero, we could have $a_3z^4+\frac{a_2}{2}z^4+\frac{a_1}{6}z^4+\frac{a_0}{24}z^4+\frac{a_{-1}}{120}z^4 \Rightarrow a_3=-\frac{1}{720}$
$\\$
Hence we have $\frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2-\frac{1}{720}z^3$.
$$\begin{align}\therefore e^z-1&=(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)-1\\&=z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots \end{align}$$
Added omitted calculation of order of pole: (thanks Chunjing Zhang for pointing it out)
$$g(z)=\frac{1}{f(z)}=\frac{1}{\frac{1}{e^z-1}}=e^z-1\\g(z_0=0)=e^0-1=0\\g'(z)=e^z \Rightarrow g(z_0=0)=e^0=1\neq 0 \\$$
Thus the order of the pole of $f(z)$ at $z_0=0$ is $1$.
Hence we let $$\frac{1}{e^z-1}=a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots$$
$$\therefore(e^z-1)(\frac{1}{e^z-1})=1\\\therefore (z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{120}+\cdots)(a_{-1}z^{-1}+a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots)=1$$
$$\Rightarrow a_{-1}+a_0z+a_1z^2+a_2z^3+\frac{1}{2}z+\frac{a_0}{2}z^2+\frac{a_1}{2}z^3+\frac{a_{-1}}{6}z^2+\frac{a_0}{6}z^3+\frac{a_{-1}}{24}z^3+\cdots=1$$
$$\therefore \begin{cases}a_{-1}=1\\a_0z+\frac{a_{-1}}{2}z=0 \Rightarrow a_0+\frac{a_{-1}}{2}=0 \Rightarrow a_0=-\frac{1}{2}\\\frac{a_{-1}}{6}z^2+\frac{a_0}{2}z^2+a_1z^2=0 \Rightarrow \frac{a_{-1}}{6}+\frac{a_0}{2}+a_1=0 \Rightarrow a_1=\frac{1}{12}\\ \frac{a_{-1}}{24}z^3+\frac{a_0}{6}z^3+\frac{a_1}{2}z^3+a_2z^3 \Rightarrow \frac{a_{-1}}{24}+\frac{a_0}{6}+\frac{a_1}{2}+a_2=0 \Rightarrow a_2=0\end{cases}$$
Therefore, the first four terms of the Laurent series:
$$\require{cancel} \cancel{1+\frac{1}{z}+(-\frac{1}{2})z^2+(0)z^3} \\ \text{Typo correction: } \frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2$$
$$\\$$
$$\\$$
The residue of given function at $z_0$ is the coefficient of $(z-z_0)^{-1}$, which is $1$.
$$\\$$
$$\\$$
If $0(z^3)=0$ is not counted since its zero, we could have $a_3z^4+\frac{a_2}{2}z^4+\frac{a_1}{6}z^4+\frac{a_0}{24}z^4+\frac{a_{-1}}{120}z^4 \Rightarrow a_3=-\frac{1}{720}$
$\\$
Hence we have $\frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+(0)z^2-\frac{1}{720}z^3$.
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Quiz-5 / Re: Q5 TUT 0203
« on: November 02, 2018, 03:59:37 PM »
We know $$\begin{align}Log(1-z)&=-\sum_{n=1}^{\infty}\frac{z^n}{n}=-(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)\\ \Rightarrow [Log(1-z)]^2&=[-(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)]^2 \\&=(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)^2 \\ &=(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)\\ &=0+(0)z+(1)z^2+(1)z^3+\frac{11}{12}z^4+\frac{5}{6}z^5+\cdots\end{align}$$
Series is valid at $|z|<1$.
Series is valid at $|z|<1$.
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MAT334--Misc / Re: Cheating during the quiz
« on: November 02, 2018, 10:25:55 AM »
Your second assumption sounds like some serious accusations against the TAs (with/without actual proof)...
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Quiz-4 / Re: Q4 TUT 0301
« on: October 26, 2018, 05:54:48 PM »
$|z+1|=2 \text{ is the circle at point -1 with radius }2.$
$$\int_{|z+1|=2}\frac{z^2}{4-z^2}dz=\int_{|z+1|=2}\frac{z^2}{(2-z)(2+z)}dz$$
Let $p=2,q=-2 \text{ where } p \text{ lies outside of the circle |z+1|=2}.$
$$\begin{align}f(z)=\frac{1}{2\pi i}\int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi \\ \Rightarrow \int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi&=2\pi i f(z)\end{align}$$
$\text{Where } f(\xi)=\frac{\xi^2}{2-\xi}, \int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi=\int_{|z+1|=2}\frac{\frac{\xi^2}{2-\xi}}{\xi-(-2)}d\xi.$ $\\$ $ f(z=-2)=\frac{(-2)^2}{2-(-2)}=1.$ $\\$
Therefore, $$ \int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi=2\pi i f(z)=2\pi i\cdot 1=2\pi i$$
$$\int_{|z+1|=2}\frac{z^2}{4-z^2}dz=\int_{|z+1|=2}\frac{z^2}{(2-z)(2+z)}dz$$
Let $p=2,q=-2 \text{ where } p \text{ lies outside of the circle |z+1|=2}.$
$$\begin{align}f(z)=\frac{1}{2\pi i}\int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi \\ \Rightarrow \int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi&=2\pi i f(z)\end{align}$$
$\text{Where } f(\xi)=\frac{\xi^2}{2-\xi}, \int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi=\int_{|z+1|=2}\frac{\frac{\xi^2}{2-\xi}}{\xi-(-2)}d\xi.$ $\\$ $ f(z=-2)=\frac{(-2)^2}{2-(-2)}=1.$ $\\$
Therefore, $$ \int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi=2\pi i f(z)=2\pi i\cdot 1=2\pi i$$
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Term Test 1 / Re: TT1 Problem 4 (morning)
« on: October 21, 2018, 04:52:46 PM »
For $$\gamma_{1}(t)=3i+3e^{it}, \gamma_{1}(t) \text{ is negatively ortiented.}$$
Thus $$\begin{align}\int_{\gamma_{1}}(z+\bar z)dz&=\int_\frac{\pi}{2}^0(\gamma_{1}(t)+\overline {\gamma_{1}(t)})\gamma_{1}'(t)dt \\ &=\frac{9}{2}e^{2it}+9it \Bigg|_{\frac{\pi}{2}}^0 \\&=9-\frac{9}{2}\pi i\end{align}$$
For $$\gamma_2(t)=3+3i-3it=3+i(3-3t), 0\leq t\leq 1.$$
$$\begin{align}\int_{\gamma_{2}}(z+\bar z)dz&=\int_0^1 (\gamma_{2}(t)+\overline {\gamma_{2}(t)})\gamma_{2}'(t)dt \\&=\int_0^1 ((3+i(3-3it))+(3-i(3-3t)))(-3i)dt \\&=\int_0^1 -18idt \\&=-18it \Big|_0^1 \\&=-18i\end{align}$$
Therefore, $$\int_L(z+\bar z)dz=\int_{\gamma_{1}}(z+\bar z)dz+\int_{\gamma_{2}}(z+\bar z)dz=9-\frac{9}{2}\pi i -18i$$
Thus $$\begin{align}\int_{\gamma_{1}}(z+\bar z)dz&=\int_\frac{\pi}{2}^0(\gamma_{1}(t)+\overline {\gamma_{1}(t)})\gamma_{1}'(t)dt \\ &=\frac{9}{2}e^{2it}+9it \Bigg|_{\frac{\pi}{2}}^0 \\&=9-\frac{9}{2}\pi i\end{align}$$
For $$\gamma_2(t)=3+3i-3it=3+i(3-3t), 0\leq t\leq 1.$$
$$\begin{align}\int_{\gamma_{2}}(z+\bar z)dz&=\int_0^1 (\gamma_{2}(t)+\overline {\gamma_{2}(t)})\gamma_{2}'(t)dt \\&=\int_0^1 ((3+i(3-3it))+(3-i(3-3t)))(-3i)dt \\&=\int_0^1 -18idt \\&=-18it \Big|_0^1 \\&=-18i\end{align}$$
Therefore, $$\int_L(z+\bar z)dz=\int_{\gamma_{1}}(z+\bar z)dz+\int_{\gamma_{2}}(z+\bar z)dz=9-\frac{9}{2}\pi i -18i$$
14
Term Test 1 / Re: TT1 Problem 3 (noon)
« on: October 20, 2018, 10:45:12 PM »
Since $(c)$ gives $u(x,y)+iv(x,y)$, the CR equation is definitely $\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}; \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$.
$$\\$$
$(b).$
$$-\frac{\partial v}{\partial x}=-(e^x\sin(y)+xe^x\sin(y)+ye^x\cos(y))=\frac{\partial u}{\partial y}$$
$$\begin{align}\Rightarrow u(x,y)&=\int (-e^x\sin(y)-xe^x\sin(y)-ye^x\cos(y))dy \\&=
xe^x\cos(y)-ye^x\sin(y)+h(x)\end{align}$$
Hence,
$$\begin{align}\frac{\partial u}{\partial x}&=e^x\cos(y)+xe^x\cos(y)-ye^x\sin(y)+h'(x)\\&=\frac{\partial v}{\partial y}=xe^x\cos(y)+e^x\cos(y)-ye^x\sin(y)\end{align}$$
$$\Rightarrow h'(x)=0\\\Rightarrow h(x)=C$$
where $C$ is an arbitrary real constant.$\\$
Therefore, $$u(x,y)=xe^x\cos(y)-ye^x\sin(y)+C$$
$$\\$$
$(c).$
$$\\$$
$$\begin{align}u(x,y)+iv(x,y)&=xe^x\cos(y)-ye^x\sin(y)+C+i(xe^x \sin (y) +y e^x\cos(y))\\&=xe^x\cos(y)+ixe^x\sin(y)+iye^x\cos(y)-ye^x\sin(y)+C\\&=xe^{x+iy}+iye^{x+iy}+C\\&=e^{x+iy}(x+iy)+C\end{align}$$
Therefore, $$\begin{align}f(z)&=ze^z+C\end{align}$$
$$\\$$
$(b).$
$$-\frac{\partial v}{\partial x}=-(e^x\sin(y)+xe^x\sin(y)+ye^x\cos(y))=\frac{\partial u}{\partial y}$$
$$\begin{align}\Rightarrow u(x,y)&=\int (-e^x\sin(y)-xe^x\sin(y)-ye^x\cos(y))dy \\&=
xe^x\cos(y)-ye^x\sin(y)+h(x)\end{align}$$
Hence,
$$\begin{align}\frac{\partial u}{\partial x}&=e^x\cos(y)+xe^x\cos(y)-ye^x\sin(y)+h'(x)\\&=\frac{\partial v}{\partial y}=xe^x\cos(y)+e^x\cos(y)-ye^x\sin(y)\end{align}$$
$$\Rightarrow h'(x)=0\\\Rightarrow h(x)=C$$
where $C$ is an arbitrary real constant.$\\$
Therefore, $$u(x,y)=xe^x\cos(y)-ye^x\sin(y)+C$$
$$\\$$
$(c).$
$$\\$$
$$\begin{align}u(x,y)+iv(x,y)&=xe^x\cos(y)-ye^x\sin(y)+C+i(xe^x \sin (y) +y e^x\cos(y))\\&=xe^x\cos(y)+ixe^x\sin(y)+iye^x\cos(y)-ye^x\sin(y)+C\\&=xe^{x+iy}+iye^{x+iy}+C\\&=e^{x+iy}(x+iy)+C\end{align}$$
Therefore, $$\begin{align}f(z)&=ze^z+C\end{align}$$