= Solution
Write $z=1+iy$ with $y\in\mathbb R$. In the <complex plane>, this is the vertical line through $1$. The <complex exponential> gives $e^z=e(\cos y+i\sin y)$, so its <complex modulus> is always $e$. Conversely, every point of that modulus has an <complex argument> $y$ and is obtained in this way. \b[The image is the entire circle centered at zero with radius $e$], traversed repeatedly as $y$ varies; it is not a disk.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/ia/paper-1-complex-exponential.png]
{title=The vertical line with real part one maps under the complex exponential to the circle of radius e}
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