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QUESTION
\begin{description}

\item[(a)]
If $z=x+jy$, where $x$ and $y$ are real numbers, express the
complex number $w=\frac{1}{1+z}$ in the form $u+jv$, where $u$ and
$v$ are real numbers.

Verify that $\ds\frac{\partial u}{\partial x}=\frac{\partial
v}{\partial y}$.

\item[(b)]
Find the general solution to the differential equation

$$\frac{dx}{dt}=e^{2x}t^2\cos(2t).$$

\end{description}


\bigskip

ANSWER

\begin{description}

\item[(a)]

\begin{eqnarray*}
w&=&\frac{1}{1+z}\\ &=&\frac{1}{1+x+jy}\\
&=&\frac{1}{(1+x+jy)}\left(\frac{1+x-jy}{1+x-jy}\right)\\
&=&\frac{1+x-jy}{(1+x)^2-j^2y^2}\\ &=&\frac{1+x-jy}{(1+x)^2+y^2}\\
&=&u+jv
\end{eqnarray*}

$$u=\frac{1+x}{(1+x)^2+y^2},\hspace{1cm}v=\frac{-y}{(1+x)^2+y^2}$$
$$\frac{\partial u}{\partial x}=\frac{
\{(1+x)^2+y^2\}(1)-(1+x)\{2(1+x)\}}{\{(1+x)^2+y^2\}^2}=
\frac{y^2-(1+x)^2}{\{(1+x)^2+y^2\}^2}$$ $$\frac{\partial
v}{\partial
y}=\frac{\{(1+x)^2+y^2\}(-1)-(-y)(2y)}{\{(1+x)^2+y^2\}^2}
=\frac{y^2-(1+x)^2}{\{(1+x)^2+y^2\}^2}$$ thus $\ds\frac{\partial
u}{\partial x}=\frac{\partial v}{\partial y}$

\item[(b)]
$\ds\frac{dx}{dt}=e^{2x}t^2\cos(2t)\\ \int e^{-2x}\,dx=\int
t^2\cos(2t)\,dt.$

Therefore
\begin{eqnarray*}
\frac{e^{-2x}}{-2}&=&
t^2\frac{\sin(2t)}{2}-\int\frac{\sin(2t)}{2}(2t)\,dt\\
&=&\frac{1}{2}t^2\sin(2t)-\left\{t\left(-\frac{\cos(2t)}{2}\right)
-\int-\frac{\cos(2t)}{2}.1\,dt\right\}\\
&=&\frac{1}{2}t^2\sin(2t)+\frac{1}{2}t\cos(2t)-\frac{1}{2}\frac{\sin(2t)}{2}+c\\
&=&\frac{1}{2}t^2\sin(2t)+\frac{1}{2}t\cos(2t)-\frac{1}{4}\sin(2t)+c\\
e^{-2x}&=&-t^2\sin(2t)-t\cos(2t)+\frac{1}{2}\sin(2t)+c\\
-2x&=&\ln\left\{\left(\frac{1}{2}-t^2\right)\sin(2t)-t\cos(2t)+c\right\}\\x&=&\ln\left\{\left(\frac{1}{2}-t^2\right)\sin(2t)-t\cos(2t)+c\right\}^{-\frac{1}{2}}
\end{eqnarray*}

\end{description}



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