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{\bf Question}

If $z=x+iy,\ w=u+iv$, and $z=w^3$ calculate $x$ and $y$ as
functions of $u$ and $v$. Substitute these expressions into the
two functions of question 1. Hence confirm that those functions
are harmonic in the $w$-plane, under the transformation $z=w^3$.

\medskip

{\bf Answer}

If $x=w^3$ then

$x+iy=(u+iv)^3=(u^3-3uv^2)+i(3u^2v-v^3)$

Therefore $\left\{\begin{array} {rcl} x & = & u^3-3uv^2\\ y & = &
3u^2v-v^3 \end{array} \right.$

Hence

\begin{description}
\item[(a)]
\begin{eqnarray*} \phi & = &
(u^3-3uv^2)^2-(3u^2v-v^3)^2+2(3u^2v-v^3)\\ & = &
u^6-15u^4v^2+15u^2v^4-v^6+6u^2v-2v^3 \end{eqnarray*}

Therefore

$\phi_{uu}=\ 30u^4-180u^2v^2+30v^4+12v$

$\phi_{vv}=-30u^4+180u^2v^2-30v^4-12v$

$\Rightarrow \nabla ^2 \phi=0$ in $w$-plane.

\item[(b)]
Must show $\phi=\sin(u^3-3uv^2) \times \cos(3u^2v-v^3)$

satisfies $\nabla ^2 \phi=0$: tedious and boring, but can do.
\end{description}

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