MAT334-2018F > Term Test 1

TT1 Problem 3 (noon)

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Victor Ivrii:
Since (c) says that $u+iv$ must be analytic,
$$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}; \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$$

Also $\bar{z}e^{z}$ is neither analytic, nor anti-analytic (satisfying "adjoint" CR)

Meng Wu:
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, $$\begin{align}f(z)&=ze^z+C\end{align}$$

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