Let $E(n,z)=|z+1|+|z^{2n}+1|$. Find $\displaystyle\min_{z\in\mathbb{C}} E(n,z)$.

Question

Let $E(n,z)=|z+1|+|z^{2n}+1|$. Find $\displaystyle\min_{z\in\mathbb{C}} E(n,z)$.

Here is what I know so far: My suspicion is that the minimum is attained for $z=\cos((2n-1)\pi/2n)+i\sin((2n-1)\pi/2n)$ where $E(n,z)=2\sin(\pi/4n)$. I manged to show this for $|z|\geq 1$; so for the question you can freely assume you are inside the unit disk.

Answer 1

Now first notice that using $z \to -z$ we need to find the minimum of $|z-1|+|z^{2n}+1|$ and then let $w=z-1$ so we need the minimum of $a(w)=|w|+|f(w)|$ where $f(w)=1+(w-1)^{2N}$.

For $w=0, z=1$ we get $2$ and for $f(w)=0, w=1-\omega_k, \omega_k^{2N}=-1, k=1,..2N$ a simple computation shows that half the time we get the OP's proposed answer $2\sin(\pi/4N)$ and half the time $2\cos(\pi/4N)$ which is manifestly bigger (for $N \ge 2$) or same for $N=1$ but always strictly less than $2$. Also by continuity any possible minimum would have $|w| \ge \delta >0$ since at zero we are away from it.

Assume now the minimum is attained at some $w \ne 0, f(w) \ne 0$. Writing $a(w)=r+R, w=re^{i\theta}, f(w)=Re^{i\phi}$ we need $\frac{\partial (r+R)}{\partial r}=0, \frac{\partial (r+R)}{\partial \theta}=0$

Since $R \ne 0$ we can get a local analytic logarithm, so $\log f =\log R +i\Phi$ and taking partial derivatives we easily see that $\frac{\partial (r+R)}{\partial \theta}=\frac{\partial R}{\partial \theta} =-R\Im \frac{wf'(w)}{f(w)}$ and $\frac{\partial (r+R)}{\partial r}=1+\frac{R}{r}\Re\frac{wf'(w)}{f(w)}$ so we get that at a minimum where $w \ne 0, f(w) \ne 0$ we must have $\frac{wf'(w)}{f(w)}$ is real and equal to $-\frac{|w|}{|f(w)|}$. Taking absolute values we get $|f'(w)|=1$ or $|1-w|^{2N-1}=\frac{1}{2N}$.

In particular $|1-w|=c_N=\frac{1}{2N}^{\frac{1}{2N-1}}<1$ and $|1-w|^{2N}=\frac{c_N}{2N}$. But then for $N \ge 3$ we already have $|f(w)| \ge 1- \frac{1}{2N} > \frac{\pi}{2N} > 2\sin \frac{\pi}{4N}$ so we cannot have a global minimum there.

For $N=2$ we use that $c_2=(\frac{1}{4})^{\frac{1}{3}}, c_2+\pi <4$ so we are good there too.

For $N=1$ we have $|w-1|=\frac{1}{2}$ and $2w(w-1)/(w^2-2w+2)=-c, c>0$ so $2w=2+\alpha, |\alpha|=1$ and $\frac{(2+\alpha)\alpha}{\alpha^2+4}$ real or $\Im(\alpha^2+2\alpha)(\bar\alpha^2+4)=0$ or $\Im(8\alpha+2\bar \alpha+4\alpha^2)=0$ or $6\sin \theta+4\sin 2\theta=0, \Im \alpha =\theta$ which gives $\cos \theta =-\frac{3}{4}$ since clearly $w-1=\pm \frac{1}{2}$ doesn't work (result is larger than $\sqrt 2$); hence $\alpha=\frac{-3 \pm i\sqrt 7}{4}, w=\frac{5 \pm i\sqrt 7}{8}$. $\Re (1+(w-1)^2) >1$ and the result is greater than $\sqrt 2$ so we are finally done!