I am trying to find the maximum value of $\left|e^{i\theta}-2\right|+\left|e^{i\theta}+2\right|\mbox{ for }0\le\theta\le2\pi$.
I can replace $e^{i\theta}$ with $a+ib$ and then proceed to get a function of $a$; Namely;
$$\sqrt{5+4a}+\sqrt{5-4a}$$ I can then use calculus to find the maximum value of this function of $a$.
My question is: Are there any more efficient ways to solve this problem?
Let $f(\theta) = \left|e^{i\theta}-2\right|+\left|e^{i\theta}+2\right|$. Note that $f(\theta)\ge0$. We have $$f^2(\theta) = |e^{i\theta}-2|^2 + |e^{i\theta}+2|^2 + 2|e^{2i\theta}-4| = (e^{i\theta}- 2)(e^{-i\theta}- 2) + (e^{i\theta}+2)(e^{-i\theta}+2) + 2|e^{2i\theta}-4| = 1 -4\cos(\theta) +4 +1+4\cos(\theta)+4 +2|e^{2i\theta}-4| = 10+ 2|e^{2i\theta}-4|$$Also we know that $$|e^{2i\theta}-4|\le |e^{2i\theta}| +|4| = 5$$So we have $$f^2(\theta)\le10+2\times 5 = 20$$We can conclude that $$0\le f(\theta)\le \sqrt{20} = 2\sqrt{5}$$If we can find $\theta_0$ such that $f(\theta_0) = 2\sqrt{5}$, we are done. Indeed, we have $f(\frac{\pi}{2}) = 2\sqrt{5}$. Also, this can be seen by $$f^2(\theta) = 20 \implies |e^{2i\theta} - 4| = 5 \implies |e^{2i\theta} - 4|^2 = 25 \implies 1 - 8\cos(2\theta) + 16 = 25 \implies 8\cos(2\theta) = -8 \implies \cos(2\theta) = -1$$Which has solutions $\theta = \frac{\pi}{2},\frac{3\pi}{2}$ in $0\le \theta \le 2\pi$.
$f(x)=\sqrt{x}, f''(x)=-\frac{1}{4x^{3/2}}<0$ so $f$ is strictly concave. By Jensen's inequality,
$$f(5+4a)+f(5-4a)\le 2f\left(\frac{(5+4a)+(5-4a)}{2}\right)=2f(5)=2\sqrt{5}$$
We have equality at $a=0, \theta = \frac{\pi}{2} \text{ or } \frac{3\pi}{2}$.
