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\textbf{Problem 1:} \\\\
\text{(a) Show that } \ 3x^{4} - 18x^{2}y^{2} + 3y^{4} - 3x \text{ is a harmonic function;} \\\\
\text{(b) Find the harmonic conjugate function } v(x,y) \\\\
\text{(c) Consider } u(x,y) + iv(x,y) \ \text{ and write it as a function } f(z) \text{ of } z=x+iy
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\textbf{Solution for part (a): } \\\\
\text{Since we have:} \\\\
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\begin{gather}
\begin{aligned}
U(x,y) = 3x^{4} - 18x^{2}y^{2} + 3y^{4} - 3x \\\\
\end{aligned}
\end{gather}
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\text{then we have the following:} \\\\
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\begin{gather}
\begin{aligned}
u_x &= 12x^{3} - 36xy^{2} - 3 \\\\
u_{xx} &= 36x^{2} - 36y^{2}\\\\
u_y &= -36x^{2}y + 12y^{3} \\\\
u_{yy} &= -36x^{2} + 36y{2}
\end{aligned}
\end{gather}
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$\text{Thus we have: } \\\\$
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\begin{gather}
\begin{aligned}
\Delta u &= u_{xx} + u_{yy} \\\\
&= 36x^{2} - 36y^{2} - 36x^{2} + 36y^{2} \\\\
&= 0
\end{aligned}
\end{gather}
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$\text{Therefore the function is harmonic. } \\\\$
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\textbf{Solution for part (b): }
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\text{We have:} \\\\
\begin{gather}
\begin{aligned}
v_x &= -u_y = 36x^{2}y - 12y^{3} \\\\
v_y &= u_x = 12x^{3} -36xy^{2} -3
\end{aligned}
\end{gather}
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\text{Then} \\\\
\begin{gather}
\begin{aligned}
v_{x,y} &= \int{v_x} \,dx + \phi(y) \\\\
&= 12x^{3}y-12xy^{3}+\phi(y)
\end{aligned}
\end{gather}
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\text{Now we calculate } \phi(y) \text{ by following:} \\\\
\begin{gather}
\begin{aligned}
v_y &= (v_{x,y})' \\\\
&= 12x^{3} -36xy^{2}-3
\end{aligned}
\end{gather}
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\text{thus, } \\\\
\begin{gather}
\begin{aligned}
\phi'(y) &= -3 \\\\
\phi(y) &= \int{-3}\,dy \\\\
&= -3y + c
\end{aligned}
\end{gather}
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\text{Therefore} \\\\
\begin{gather}
\begin{aligned}
v_{x,y} = 12x^{3}y-12xy^{3}-3y+c
\end{aligned}
\end{gather}
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\textbf{Solution for part (c): }
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\text{Let } \ z=x+iy \\\\
\begin{gather}
\begin{aligned}
u_{x,y} + iv_{x,y} &= 3x^{4} -12x^{2}y^{2} + 3y^{4} -3x +i(12x^{3}y-12xy^{3}-3y)+iC \\\\
&= 3(3x^{4} -12x^{2}y^{2} + 3y^{4} + i12x^{3}y-i12xy^{3}) -3(x+iy) +iC\\\\
&= 3(x+iy)^{4}-3(x+iy)+iC\\\\
&= 3z^{4} -3z + iC
\end{aligned}
\end{gather}
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