= Solution
{c}
Taking the imaginary part of the <complex potential of a two-dimensional potential flow>[complex potential] gives the <streamfunction>
$$
\boxed{\psi=\beta[(x^2-y^2)\sin\alpha+2xy\cos\alpha]
=\beta r^2\sin(2\theta+\alpha).}
$$
The corresponding <velocity field> is
$$
\begin{pmatrix}v_x\\v_y\end{pmatrix}
=2\beta
\begin{pmatrix}\cos\alpha&-\sin\alpha\\-\sin\alpha&-\cos\alpha\end{pmatrix}
\begin{pmatrix}x\\y\end{pmatrix}.
$$
Put $\zeta=z e^{i\alpha/2}=\xi+i\eta$. Then $w=\beta\zeta^2$ and $\psi=2\beta\xi\eta$. Thus the <streamlines> are the previous rectangular <hyperbolas> rotated clockwise by $\alpha/2$. Their separatrices have angles $\theta=-\alpha/2+n\pi/2$; the extension axis is $\theta=-\alpha/2$ and the compression axis is perpendicular to it. The middle panel of the figure shows this rotation.
For pure <advection>, a marked <fluid element> obeys
$$
\dot\xi=2\beta\xi,\qquad \dot\eta=-2\beta\eta,
\qquad
\xi(t)=e^{2\beta t}\xi(0),\quad \eta(t)=e^{-2\beta t}\eta(0).
$$
Consequently the dye circle becomes an <ellipse> with semiaxes $Re^{2\beta t}$ and $Re^{-2\beta t}$. Its <area> is
$$
\boxed{\mathcal A(t)=\pi(Re^{2\beta t})(Re^{-2\beta t})=\pi R^2.}
$$
Stretching in one principal direction exactly compensates compression in the other. Equivalently the flow-map <Jacobian determinant> is one, as required by <incompressibility>. Negligible <diffusion> is needed to regard the sharp dye boundary as a <material curve>.
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