Question

Consider the set of complex number $A, B, C$ and $S$ defined as

$A = \{z : ||z+2| - |z-2|| = 2\}$ $B = \left\{z : \arg\left(\frac{z-1}{z}\right) = \frac{\pi}{2}\right\}$ $C = \{z : \arg(z-1) = \pi\}$ $S = \left\{z : \operatorname{Re}\left(\frac{z-1}{z+1}\right) = 0\right\}$

If $z_1, z_2, z_3 \in S$, then find the minimum value of $|z_1 + z_2|^2 + |z_2 + z_3|^2 + |z_3 + z_1|^2$.

Answer 1

3

Answer 2

The given sets are:

$A = \{z : ||z+2| - |z-2|| = 2\}$ represents a hyperbola with foci at $(-2,0)$ and $(2,0)$ and transverse axis length $2a=2$. The equation is $x^2 - \frac{y^2}{3} = 1$.

$B = \left\{z : \arg\left(\frac{z-1}{z}\right) = \frac{\pi}{2}\right\}$ represents the upper semi-circle of the circle with diameter connecting $0$ and $1$. The equation is $(x - 1/2)^2 + y^2 = (1/2)^2$ with $y > 0$.

$C = \{z : \arg(z-1) = \pi\}$ represents the ray on the real axis from $-\infty$ to $1$, i.e., $\{z=x+i0 : x < 1\}$.

$S = \left\{z : \operatorname{Re}\left(\frac{z-1}{z+1}\right) = 0\right\}$ represents the circle with diameter connecting $1$ and $-1$, excluding the points $1$ and $-1$. The equation is $x^2+y^2=1$, excluding $z=1$ and $z=-1$. This is the unit circle centered at the origin, excluding the points $(1,0)$ and $(-1,0)$.

We are given $z_1, z_2, z_3 \in S$. This means $|z_1| = |z_2| = |z_3| = 1$. We want to find the minimum value of $L = |z_1 + z_2|^2 + |z_2 + z_3|^2 + |z_3 + z_1|^2$. Using the property $|w|^2 = w\bar{w}$ and $|z|=1 \implies \bar{z} = 1/z$, we have: $|z_i + z_j|^2 = (z_i + z_j)(\bar{z_i} + \bar{z_j}) = z_i\bar{z_i} + z_i\bar{z_j} + z_j\bar{z_i} + z_j\bar{z_j} = |z_i|^2 + z_i\bar{z_j} + z_j\bar{z_i} + |z_j|^2$. Since $|z_i|=1$ and $|z_j|=1$, this simplifies to $1 + z_i\bar{z_j} + \overline{z_i\bar{z_j}} + 1 = 2 + 2 \operatorname{Re}(z_i\bar{z_j})$.

So, $L = (2 + 2 \operatorname{Re}(z_1\bar{z_2})) + (2 + 2 \operatorname{Re}(z_2\bar{z_3})) + (2 + 2 \operatorname{Re}(z_3\bar{z_1}))$. $L = 6 + 2 (\operatorname{Re}(z_1\bar{z_2}) + \operatorname{Re}(z_2\bar{z_3}) + \operatorname{Re}(z_3\bar{z_1}))$.

Let $z_k = e^{i\theta_k}$ for $k=1, 2, 3$. Then $z_1\bar{z_2} = e^{i\theta_1}e^{-i\theta_2} = e^{i(\theta_1-\theta_2)}$, so $\operatorname{Re}(z_1\bar{z_2}) = \cos(\theta_1-\theta_2)$. Similarly, $\operatorname{Re}(z_2\bar{z_3}) = \cos(\theta_2-\theta_3)$ and $\operatorname{Re}(z_3\bar{z_1}) = \cos(\theta_3-\theta_1)$.

We want to minimize $L = 6 + 2 (\cos(\theta_1-\theta_2) + \cos(\theta_2-\theta_3) + \cos(\theta_3-\theta_1))$. Let $\alpha = \theta_1-\theta_2$, $\beta = \theta_2-\theta_3$, $\gamma = \theta_3-\theta_1$. Note that $\alpha+\beta+\gamma = 0$. We want to minimize $\cos(\alpha) + \cos(\beta) + \cos(\gamma)$ subject to $\alpha+\beta+\gamma=0$. The sum $\cos(\alpha) + \cos(\beta) + \cos(\gamma)$ is minimized when $\alpha, \beta, \gamma$ are as spread out as possible, subject to their sum being 0. This typically occurs when $\alpha = \beta = \gamma = 0$ (maximum sum 3) or when they are angles of an equilateral triangle, i.e., $\alpha = \beta = \gamma = 2\pi/3$ or $-2\pi/3$ (sum -3/2) or when two angles are $\pi$ and one is $-2\pi$ (sum -1). Consider the case when $z_1, z_2, z_3$ form an equilateral triangle inscribed in the unit circle. Let $z_1 = 1 = e^{i0}$, $z_2 = e^{i2\pi/3}$, $z_3 = e^{i4\pi/3}$. These points are in $S$ since they are on the unit circle and are not $1$ or $-1$. In this case, $\theta_1 = 0$, $\theta_2 = 2\pi/3$, $\theta_3 = 4\pi/3$. $\theta_1 - \theta_2 = -2\pi/3$. $\cos(-2\pi/3) = -1/2$. $\theta_2 - \theta_3 = 2\pi/3 - 4\pi/3 = -2\pi/3$. $\cos(-2\pi/3) = -1/2$. $\theta_3 - \theta_1 = 4\pi/3 - 0 = 4\pi/3$. $\cos(4\pi/3) = -1/2$. The sum $\cos(\theta_1-\theta_2) + \cos(\theta_2-\theta_3) + \cos(\theta_3-\theta_1) = -1/2 + (-1/2) + (-1/2) = -3/2$.

The minimum value of $\cos(\alpha) + \cos(\beta) + \cos(\gamma)$ subject to $\alpha+\beta+\gamma=0$ is $-3/2$. This occurs when $\alpha=\beta=\gamma = \pm 2\pi/3$. The minimum value of $L$ is $6 + 2(-3/2) = 6 - 3 = 3$.

This minimum is achieved when $z_1, z_2, z_3$ represent the vertices of an equilateral triangle inscribed in the unit circle. For example, $z_1=1, z_2=e^{i2\pi/3}, z_3=e^{i4\pi/3}$. These points are in $S$.

The final answer is $\boxed{3}$.

Explanation:

The set $S$ is the unit circle centered at the origin, excluding points $1$ and $-1$. For $z \in S$, we have $|z|=1$. The expression to minimize is $|z_1 + z_2|^2 + |z_2 + z_3|^2 + |z_3 + z_1|^2$. Using $|w|^2 = w\bar{w}$ and $\bar{z}=1/z$ for $|z|=1$, the expression simplifies to $6 + 2(\operatorname{Re}(z_1\bar{z_2}) + \operatorname{Re}(z_2\bar{z_3}) + \operatorname{Re}(z_3\bar{z_1}))$. Let $z_k = e^{i\theta_k}$. The expression becomes $6 + 2(\cos(\theta_1-\theta_2) + \cos(\theta_2-\theta_3) + \cos(\theta_3-\theta_1))$. Let $\alpha = \theta_1-\theta_2$, $\beta = \theta_2-\theta_3$, $\gamma = \theta_3-\theta_1$. Then $\alpha+\beta+\gamma=0$. The minimum value of $\cos(\alpha) + \cos(\beta) + \cos(\gamma)$ subject to $\alpha+\beta+\gamma=0$ is $-3/2$, which occurs when $\alpha=\beta=\gamma=\pm 2\pi/3$. This corresponds to $z_1, z_2, z_3$ forming an equilateral triangle inscribed in the unit circle. The minimum value of the expression is $6 + 2(-3/2) = 3$.