= Solution
Writing $z=x+iy$ gives $|w|=e^x$ and $\arg w=y$. Thus the closed square maps to the annular sector
$$
\boxed{1\leq|w|\leq e^a,\qquad0\leq\arg w\leq a.}
$$
Its horizontal edges become radial segments and its vertical edges become circular arcs. Each radial tangent is perpendicular to the tangent of its adjoining circle, so the four interior angles remain $\pi/2$. Equivalently, the complex <derivative> $e^z$ is nonzero at every vertex and the map is <conformal> there.
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