Find maximun of $ f(z) = e^z - e $ in a rectangle

Question

$S$ is a rectangle with the vertices $\pm1$ and $\pm \pi i$, and $f(z) = e^z - e$. I need to find the maximum of $|f(z)|$ on $S$ and the points where it happens.

I know the maximum appears on the rectangle itself but how do I find it?

Thanks.

Answer 1

Let $z=x+iy$. Then,

$$|e^z-e|^2= (e^z-e) (e^{\bar{z}}-e)=e^{2x}-2e^{x+1}\cos y + e^2=(e^x+e)^2-4e^{x+1}\cos^2\frac y2$$

Note that, for any given $x$,

$$|e^z-e|^2 \le (e^x+e)^2$$

where the equality is at $\cos\frac y2 = 0$, or $y=\pm\pi$. Moreover, recognize that the maximum value of $e^x+e$ occurs when $x=1$. Thus, the maximum of $f(z)$ is

$$f(1\pm\pi i) = e+e = 2e$$

Answer 2

HINT

Here is one approach. Parameterize each of the four sides of $S$. For example, the first side goes from $1$ to $\pi i$, so you have $z(t) = (1-t) + \pi ti$ for $t \in [0,1]$.

Now compute $|e^{z(t)}-e|$ as a function of $t$ using Euler's Theorem, for example, and maximize over $t$ in regular 1-variable optimization.

Then do this for all 4 sides (or use symmetry to make your work shorter).