On the line segment from $z=R, R>0,$ to $z=R+i2\pi$, I want to find the maximum of: $\lvert e^{3z}/(1+e^z) \rvert$. If $z=x+iy$, this is equal to:
$$ \lvert e^{3R}e^{i3y} / (1+e^Re^{iy}) \rvert $$
This is from problem 14 b) in Saff & Snider, and in the solutions they give a very pretty answer but I can't see how they did it. Been a while since I calculated with absolute values so any tips and tricks would be helpful.
On
Maximising the absolute value is the same as maximising the square of the absolute value. Using that $\lvert w \rvert^2 =\bar{w}w $, we have $$ \left\lvert \frac{e^{3z}}{1+e^z} \right\rvert^2 = \frac{e^{3(z+\bar{z})}}{(1+e^z)(1+e^{\bar{z}})} = \frac{e^{6\Re{(z)}}}{1+2\Re(e^z)+e^{2\Re(z)}}, $$ using that $2\Re(w) = w+\bar{w}$. Now, we have $\Re(z) = R$, $e^z = e^R e^{iy}$, so $\Re(e^z) = e^R \cos{y}$. Therefore we have $$ \left\lvert \frac{e^{3z}}{1+e^z} \right\rvert^2 = \frac{e^{6R}}{1+e^{2R}+2e^{R}\cos{y}}. $$ Now you can see that this is maximised when the denominator is smallest: this occurs when $\cos{y}=-1$, or $y=\pi$. Then you find that the minimum of the absolute value is $$ \frac{e^{3R}}{\sqrt{1+e^{2R}-2e^{R}}} = \frac{e^{3R}}{\lvert e^R-1 \rvert}. $$
Hint: To maximize $|1/f|$ you want to minimize $|f|$. As $y$ goes from $0$ to $2\pi$, $e^R e^{iy}$ goes around a circle of radius $e^R$ centred at $0$. What is the closest point to $-1$ on that circle?