Showing internal angles of a square are unaffected by a mapping

Question

I recently had an exam in complex analysis, and I am slightly confused by one of the questions, so I'd appreciate any clarification:

The mapping from the complex $z$ plane to the complex $w$ plane is given by: $$f:z\mapsto w=e^{z}$$ Find the image in the $w$ complex plane of the map $f$ applied to the square in the $z$ complex plane with vertices at $0$, $a$, $a(1+i)$, and $ai$, where $0 < a < \pi/2$ is real. Show that the internal angles at the vertices of the square are unchanged by the mapping.

I am unsure how this can possibly be true. If we map all of the points to the $w$-plane using $f(z)$, then we have:

$$w_{1}=e^{0}=1,\: w_{2}=e^{a},\: w_{3} = e^{a(1+i)}=e^{a}(\cos(a)+i\sin(a)),\:w_{4} = e^{ai} = \cos(a) + i\sin(a)$$

If we plot this in the complex $w$ plane, for a reasonable value of $a$, for instance $a=\pi/4$, then we have something like the following:

We can see that the top two vertices have their angles preserved, but the bottom two do not. Is the question wrong, or am I misunderstanding something fundamental?

Answer 1

The exponential function maps the real line to the positive reals, so your bottom line is good.

Look at the mapping of the line from $0$ to $ai$ where a typical point $ci$ is mapped to $e^{ci} = \cos c +i\sin c$ which gives you not a straight line, but a circular arc of radius $1$ centred at the origin. If you draw this you will find a right-angle in the image at the point $1$.

The exponential function does not, in general, map straight lines to straight lines.

Answer 2

Your assumption about the lines between your corner points is not correct. To clarify my point you should try to calculate the images of the midpoints of every of these lines that connect the corner points. What you should do, is take a look at the derivative of your function at the points of interest, since from there you can calculate tangents and from there the angle between those tangents.