Minimum of $E(z)=|z+1|+|z^{2n}+1|$

Question

Let $n \in \mathbb{N}$. Find the minimum of the expression $E(z)=|z+1|+|z^{2n}+1|$ over $\mathbb{C}$.

I found this problem in a Romanian magazine with all sorts of math problems.
(Edit: ${\color{blue}{\textrm{This was suggested for a contest for 10th graders}}}$, According to @alexanderv's comment.)

What I've done so far(WRONG): $E \geq 0$ so if we can find $z$ with $E(z)=0$ it's enough. If $n=0$, the minimum is achieved in $-1$, $E(z) =2$. If $n \geq 1$, let's take $z= \cos \theta + i\sin \theta$. Using $|1+\cos \alpha + i\sin \alpha|=2| \cos (\frac{\alpha}{2})|$, we get $E(z)=2(|cos(\frac{\theta}{2})|+|cos(n\theta)|)$. Then, taking the modulus and using $\cos(\pi-\alpha)=-\cos(\alpha)$ and the $\cos(x) + \cos(y)$ formula, we take $z$ so one of the product terms equals to $0$, achieving the minimum. But I'm not sure if it's right, can someone approve?

What's wrong above is $E(z)=0$ implies $z=-1$, that means $E(z)=2$, contradiction. I don't have any idea how to find the minimum.

Answer 1

Solution:

Fact 1: $|z + 1| + |z^2 + 1| \ge \sqrt{2}$ for all $z\in \mathbb{C}$, with equality if $z = \mathrm{i}, -\mathrm{i}$.

Fact 2: Let $n \ge 2$ be a fixed integer. Then $$|z + 1| + |z^{2n} + 1| \ge 2\sin \frac{\pi}{4n}, \ \forall z \in \mathbb{C}$$ with equality if $z = \mathrm{e}^{\mathrm{i}\frac{2n - 1}{2n}\pi}, \ \mathrm{e}^{\mathrm{i}\frac{2n + 1}{2n}\pi}$.

$\phantom{2}$

Proof of Fact 1:

(Solution due to ali3985@AoPS) We have \begin{align*} |z + 1| + |z^2 + 1| &= |z + \mathrm{i} - (\mathrm{i} - 1)| + |z - \mathrm{i}|\cdot |z + \mathrm{i}|\\ &\ge |\mathrm{i} - 1| - |z + \mathrm{i}| + |z - \mathrm{i}|\cdot |z + \mathrm{i}|\\ &= \sqrt{2} + |z + \mathrm{i}|(|z - \mathrm{i}| - 1). \end{align*} Also, we have \begin{align*} |z + 1| + |z^2 + 1| &= |z - \mathrm{i} + (\mathrm{i} + 1)| + |z - \mathrm{i}|\cdot |z + \mathrm{i}|\\ &\ge |\mathrm{i} + 1| - |z - \mathrm{i}| + |z - \mathrm{i}|\cdot |z + \mathrm{i}|\\ &= \sqrt{2} + |z - \mathrm{i}|(|z + \mathrm{i}| - 1). \end{align*} We have $|z - \mathrm{i}| + |z + \mathrm{i}| \ge |(z - \mathrm{i}) - (z + \mathrm{i})| = 2$. Thus, either $|z - \mathrm{i}| - 1 \ge 0$ or $|z + \mathrm{i}| - 1 \ge 0$.

We are done.

$\phantom{2}$

Proof of Fact 2:

When $|z| > 1$, we have \begin{align*} |z + 1| + |z^{2n} + 1| &= |z| \cdot |1/z + 1| + |z|^{2n}\cdot |1/z^{2n} + 1|\\ &\ge |1/z + 1| + |(1/z)^{2n} + 1|. \end{align*} Thus, we only need to prove the case when $|z| \le 1$.

We split into two cases:

1. $0 \le |z| \le (3/5)^{1/n}$:

We have \begin{align*} f(z) &\ge 1 - |z| + 1 - |z|^{2n} \\ &\ge \frac{41}{25} - (3/5)^{1/n} \\ &\ge \frac{41}{25} - \left(1 - \frac{2}{5n}\right) \qquad\qquad (\textrm{Bernoulli inequality})\\ &> \frac{\pi}{2n} \\ &\ge 2\sin \frac{\pi}{4n} \end{align*} where we have used $u\ge \sin u$ for all $u \ge 0$ in the last inequality.

2. $(3/5)^{1/n} < |z| \le 1$:

Let $z = r \mathrm{e}^{\mathrm{i}\theta}$ ($(3/5)^{1/n} < r \le 1$, $\theta \in [0, 2\pi]$). We have \begin{align*} f(z) &= \sqrt{r^2 + 1 + 2r \cos \theta} + \sqrt{r^{4n} + 1 + 2r^{2n} \cos 2n\theta}\\ &= \sqrt{(1 - r)^2 + 2r(1 + \cos \theta)} + \sqrt{(1 - r^{2n})^2 + 2r^{2n}(1 + \cos 2n\theta)} \\ &\ge \sqrt{2r(1 + \cos \theta)} + \frac{1}{\sqrt{2}} \left(1 - r^{2n} + \sqrt{2}\, r^n\sqrt{1 + \cos 2n\theta}\right) \tag{1}\\ &= \sqrt{2r(1 + \cos \theta)} + \frac{1 - r^{2n}}{\sqrt{2}} + r^n\sqrt{1 + \cos 2n\theta}\\ &\ge \sqrt{2r(1 + \cos \theta)} + \frac{3(1 - \sqrt r)}{\sqrt{2}} + \frac35 \sqrt{1 + \cos 2n\theta} \tag{2}\\ &= - \left(\frac{3}{\sqrt{2}} - \sqrt{2(1 + \cos \theta)}\right)\sqrt{r} + \frac{3}{\sqrt{2}} + \frac{3}{5} \cdot\sqrt{1 + \cos 2n\theta}\\ &\ge - \left(\frac{3}{\sqrt{2}} - \sqrt{2(1 + \cos \theta)}\right) + \frac{3}{\sqrt{2}} + \frac{3}{5} \cdot\sqrt{1 + \cos 2n\theta} \tag{3}\\ &= \sqrt{2(1 + \cos \theta)} + \frac{3}{5} \cdot\sqrt{1 + \cos 2n\theta}\\ &= 2|\cos (\theta/2)| + \frac{3}{5}\sqrt{2} |\cos n\theta|\\ &\ge 2|\cos (\theta/2)| + \frac{4}{5} |\cos n\theta|. \end{align*} Explanations:
(1): $\sqrt{a^2 + b^2} \ge \frac{a + b}{\sqrt 2}$.
(2): $1 - r^{2n} \ge 1 - r^2 = (1 - \sqrt{r})(1 + \sqrt{r})(1 + r)$ and $(1 + \sqrt{r})(1 + r) \ge (1 + \sqrt{(3/5)^{1/2}})(1 + (3/5)^{1/2}) > 3$; and $r^n \ge \frac35$.
(3): $\frac{3}{\sqrt{2}} - \sqrt{2(1 + \cos \theta)} \ge \frac{3}{\sqrt{2}} - \sqrt{2(1 + 1)} > 0$.

It suffices to prove that, for all $\theta\in [0, 2\pi]$, $$2|\cos (\theta/2)| + \frac{4}{5} |\cos n\theta| \ge 2\sin \frac{\pi}{4n}.$$ WLOG, assume $\theta \in [0, \pi]$. Then $2|\cos (\theta/2)| = 2\cos (\theta/2)$. We only need to prove the case when $2\cos (\theta/2) \le 2\sin \frac{\pi}{4n}$ that is $\pi - \frac{\pi}{2n} \le \theta \le \pi$. Let $\alpha = n\theta - n\pi + \pi/2 \in [0, \pi/2]$. We have $\sin \alpha = (-1)^n \cos n\theta$ and $|\cos n\theta| = \sin \alpha$. Also, we have $2\cos (\theta/2) = 2\sin(\frac{\pi}{4n} - \frac{\alpha}{2n})$.

It suffices to prove that, for all $\alpha \in [0, \pi/2]$, $$2\sin\left(\frac{\pi}{4n} - \frac{\alpha}{2n}\right) + \frac{4}{5}\sin \alpha - 2\sin \frac{\pi}{4n} \ge 0$$ that is $$\frac{4}{5}\sin \alpha \ge 4 \cos \frac{\pi - \alpha}{4n} \sin \frac{\alpha}{4n}.$$

It suffices to prove that, for all $\alpha \in [0, \pi/2]$, $$\frac{4}{5}\sin \alpha \ge 4 \sin \frac{\alpha}{8}$$ which is true by using the well-known fact $\frac{2}{\pi}u \le \sin u \le u$ for all $u \in [0, \pi/2]$.

We are done.

Answer 2

$\def\z{\bar z}$ Let $$E_m(z)=|z+1|+|z^{m}+1|;\quad m\in2\mathbb Z_+.\tag1 $$ It is convenient to rewrite the equation $(1)$ as: $$ E_m(z,\z)=\sqrt{(z+1)(\z+1)}+\sqrt{(z^m+1)(\z^m+1)},\tag2 $$ where $\z$ is the complex conjugate of $z$, and treat $z$ and $\z$ as independent variables (they are of course not independent but this is a well-known and very convenient tool for finding the critical points). Obviously the function $E_m(z,\z)$ is differentiable at all points $(z,\z)$ unless one of the square roots is $0$.

As easy to check among $m$ solutions of the equation $z^{m}+1=0$ the root $z=e^{i\frac{m-1}{m}\pi}$ delivers the least value of $E_m$ which reduces to $$|z+1|=2\sin\tfrac\pi{2m}.\tag3$$

We claim that $(3)$ is the global minimum of $E_m(z)$. To prove this it suffices to show that the value of the function $E_m(z)$ at all stationary points is higher than that given by $(3)$.

The stationary points can be determined from the equations: $$\begin{align} &\frac{\partial E_m}{\partial z}=0 \implies \frac{(\z^m+1)m z^{m-1}}{\sqrt{(z^m+1)(\z^m+1)}}= -\frac{\z+1}{\sqrt{(z+1)(\z+1)}};\tag{4a}\\ &\frac{\partial E_m}{\partial \z}=0 \implies \frac{(z^m+1)m \z^{m-1}}{\sqrt{(z^m+1)(\z^m+1)}}= -\frac{z+1}{\sqrt{(z+1)(\z+1)}}.\tag{4b}\\ \end{align}$$

Multiplying the corresponding sides of the equations one obtains: $$ m^2|z|^{2(m-1)}=1\implies |z|=m^{-\frac1{m-1}}\equiv\rho_m.\tag5 $$

The equation $(5)$ means that the stationary points $z_s$ and their images $z_s^m$ lie on the concentric circles with center at $(0,0)$ and radii $\rho_m$ and $\rho_m^{m}$, respectively ($\rho_m^{m}<\rho_m<1$). This finding helps to visualize the problem since the function $E_m(z_s)$ is geometrically the sum of distances from the point $z=-1$ to $z_s$ and $z_s^m$, which is bounded from below by the value: $$ (1-\rho_m)+(1-\rho^{m}_m)=2-\left(1+\frac1{m}\right)\rho_m>1.\tag6 $$ The last inequality for $m>1$ follows most simply from $x\log(x)+(1-x)\log(1+x)<0$ valid for all $x:\ 0<x<1$.

Since $2\sin\frac\pi{4n}<1$ for $n\ge2$ it remains to consider only the case $n=1$. For this we explicitly find the solutions of $(4)$ for $m=2$. To facilitate the computation we substitute $z=\rho_2e^{i\phi}=\tfrac12e^{i\phi}$ in one of the equations. After straightforward algebra one finds the solutions $z_1=-\tfrac12$, $z_{2,3}=\tfrac{-3\pm i\sqrt7}8$ and determines: $$E_2(z_{2,3})=\tfrac54\sqrt2> E_2(z_1)=\tfrac74>2\sin\tfrac\pi4=\sqrt2,$$ which finalizes the proof.

Answer 3

[Update 09 April 2021] Here is a partial proof for the case $|z| \ge 1$.

We have with $z = r e^{-i \phi}$: $$\begin{align} E(z)=&|z+1|+|z^{2n}+1| = |r e^{-i \phi}+1|+|r^{2n} e^{-i 2 n \phi}+1| \\ =& |r + e^{i \phi}|+|r^{2n} + e^{i 2 n \phi}| \\ =& \sqrt{(r + \cos\phi)^2 + \sin^2\phi} + \sqrt{(r^{2n} + \cos(2n\phi))^2 + \sin^2(2n\phi)}\\ =& \sqrt{r^2 - 2 r +1 + 2 r (1+\cos\phi)} + \sqrt{r^{4n} - 2 r^{2n} +1 + 2 r^{2n} (1+\cos(2n \phi))} \\ =& \sqrt{(r-1)^2 + 4 r \cos^2(\phi/2)} + \sqrt{(r^{2n} -1)^2 + 4 r^{2n} \cos^2( n \phi)} \\ \ge& 2 \sqrt{ r} |\cos(\phi/2)| + 2 r^n |\cos(n \phi)| \qquad {\rm\bf [1]}\\ \ge& 2 |\cos(\phi/2)| + 2 |\cos( n \phi)| \qquad {\rm\bf [2]} \end{align}$$

Note in ${\rm\bf [1]}$ that this step is tight for $r=1$ and in ${\rm\bf [2]}$ that both $\sqrt{r}$ and $r^n$ are rising functions, so, whatever $\phi$, the smallest value is given for $r=1$ (in the considered range $r \ge 1$), so this is again tight. So the last result is indeed the correct minimum for $r \ge 1$.

The last result has now to be minimized w.r.t. $\phi$. W.l.og. we can take $0\le \phi\le \pi$ so we have to minimize $2 \cos(\phi/2) + 2 |\cos( n \phi)|$. Since $\cos(\phi/2)$ is falling for $0\le \phi\le \pi$, the minimum will be attained closest to $\pi$ at a point where $\cos( n \phi)$ changes sign, i.e. at $\phi = (\frac12 + k) \pi/n $ with the integer $k$ chosen such that $\phi$ comes closest to $\pi$, i.e. $k = n-1$. This gives $\phi = \pi \frac{2n-1}{2n}$ and hence the minimum for $|z| \ge 1$ is obtained as
$$E(z) \ge 2 \cos(\pi \frac{2n-1}{4n}) = 2 \sin(\frac{\pi}{4n}) $$

What remains to be shown is that for $|z| <1$, no smaller values of $E(z)$ are attained. The above method does not work, as the step ${\rm\bf [1]}$ above neglects the terms $(r-1)^2$ and $(r^{2n}-1)^2$ which are falling for $r<1$. Hence as $r$ moves away from $1$ (i.e. gets smaller), the true value of $E(z)$ is increased by these terms, so the bound ${\rm\bf [1]}$ used above gives way too small values for $r<1$. However, numerical evidence indeed supports that the minimum of $E(z)$ is attained at $|z| =1$, so in this case the derived minimum is indeed valid for all $z$.

I didn't succeed to do that part of the proof yet, suggestions are welcome.

Answer 4

shymilan@AoPS's elegant proof (I rewrote it):

Let $w_k = \mathrm{e}^{\mathrm{i}\frac{(2k - 1)}{2n}\pi}, ~ k = 1, 2, \cdots, 2n$. Then $Q(z) := z^{2n} + 1 = (z - w_1)(z - w_2)\cdots (z - w_{2n})$. We have the partial fraction decomposition $$\frac{1}{z^{2n} + 1} = \frac{1}{Q(z)} = \sum_{k=1}^{2n} \frac{1}{Q'(w_k)}\frac{1}{z - w_k} = \sum_{k=1}^{2n} \frac{-w_k}{2n}\frac{1}{z - w_k}. \tag{1}$$ (easy to prove, e.g. see Partial_fraction_decomposition).

For $k = 1, 2, \cdots, 2n$, we have \begin{align*} |z + 1| + |z^{2n} + 1| &= |z - w_k + w_k + 1| + |z-w_k| \cdot \left|\frac{z^{2n} + 1}{z - w_k}\right|\\ &\ge |w_k + 1| - |z - w_k| + |z-w_k| \cdot \left|\frac{z^{2n} + 1}{z - w_k}\right|\\ &= |w_k + 1| + |z - w_k| \left(\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1\right). \tag{2} \end{align*}

Using (1), we have \begin{align*} \sum_{k=1}^{2n} \left(\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1\right) &= \sum_{k=1}^{2n} \left|\frac{z^{2n} + 1}{z - w_k}\right| - 2n \\ &= \sum_{k=1}^{2n} \left|\frac{z^{2n} + 1}{z - w_k} \cdot w_k\right| - 2n\\ &\ge \left|\sum_{k=1}^{2n} \frac{z^{2n} + 1}{z - w_k} \cdot w_k\right| - 2n\\ &= 0. \tag{3} \end{align*}

From (2) and (3), there exists $k \in \{1, 2, \cdots, 2n\}$ such that $\left|\frac{z^{2n} + 1}{z - w_k}\right| - 1 \ge 0$ and $$|z + 1| + |z^{2n} + 1| \ge |w_k + 1|.$$

For $k = 1, 2, \cdots, 2n$, we have $$|w_k + 1| = |\mathrm{e}^{\mathrm{i}\frac{(2k - 1)}{2n}\pi} + 1| = 2\left|\sin \frac{(2n - 2k + 1)\pi}{4n}\right| \ge 2 \sin \frac{\pi}{4n}.$$

Thus, we have $$|z + 1| + |z^{2n} + 1| \ge 2 \sin \frac{\pi}{4n}.$$

Also, if $z = \mathrm{e}^{\mathrm{i}\frac{2n - 1}{2n}\pi}$, we have $|z+1| + |z^{2n} + 1| = 2 \sin \frac{\pi}{4n}$.

Thus, the minimum of $|z+1| + |z^{2n} + 1|$ is $2 \sin \frac{\pi}{4n}$.