Problem 1:
(a)Show that
\begin{align*} u(x,y)=sin(4x)e^{-4y}+3xy.\end{align*}
is a harmonic function.
(b) Find a harmonic conjugate function $v(x,y)$.
(c)Consider $u(x,y)+iv(x,y)$ and write it as a function $f(z)$ of $z=x+iy$.
Answer:
(a)
\begin{align*}
u(x,y)&=sin(4x)e^{-4y}+3xy\\
u_x &= 4cos(4x)e^{-4y}+3y\\
u_{xx} &= -16sin(4x)e^{-4y}\\
u_y &= -4sin(4x)e^{-4y}+3x\\
u_{yy} &= 16sin(4x)e^{-4y}\\
\end{align*}
Then,
\begin{align*}
\Delta u &= u_{xx}+u_{yy}\\
&=-16sin(4x)e^{-4y}+16sin(4x)e^{-4y}\\
&=0
\end{align*}
Thus, $u(x,y)$ is a harmonic function.
(b) We know
\begin{cases}
v_x = -u_y = 4sin(4x)e^{-4y}-3x\\
v_y = u_x = 4cos(4x)e^{-4y}+3y
\end{cases}
Then,
\begin{align*}
v(x,y) &= \int v_x\ dx\\
&=\int 4sin(4x)e^{-4y}-3x\ dx\\
&=-cos(4x)e^{-4y}- \frac{3}{2}x^2+\phi (y)\\
v_y &= 4cos(4x)e^{-4y} + \phi '(y)\\
&= 4cos(4x)e^{-4y}+3y\\
\end{align*}
Now, we have
\begin{align*}
\phi '(y) &= 3y\\
\phi (y) &= \frac{3}{2}y^2 +C\\
v(x,y) &= -cos(4x)e^{-4y}- \frac{3}{2}x^2+\frac{3}{2}y^2 +C\\
\end{align*}
(c) Consider $u(x,y)+iv(x,y)$, we have
\begin{align*}
u(x,y)+iv(x,y) &= sin(4x)e^{-4y}+3xy + i(-cos(4x)e^{-4y}- \frac{3}{2}x^2+\frac{3}{2}y^2 +C)\\
&= sin(4x)e^{-4y}+3xy -icos(4x)e^{-4y}- i\frac{3}{2}x^2+i\frac{3}{2}y^2 +iC\\
&=e^{-4y}(sin(4x)-icos(4x))-i\frac{3}{2}(x^2-y^2+2ixy)+iC\\
&= e^{-4y}(-i)(cos(4x)+isin(4x))-i\frac{3}{2}(x+iy)^2+iC\\
&=-ie^{-4y+i4x}-i\frac{3}{2}(x+iy)^2+iC\\
&=-ie^{4i(iy+x)}-i\frac{3}{2}(x+iy)^2+iC\\
&= -ie^{4iz}-i\frac{3}{2}z^2+iC\ \ \ \text{for $z=x+iy$}\\
\end{align*}
Therefore, $f(x) = -ie^{4iz}-i\frac{3}{2}z^2+iC$
