Let $z_1, z_2, z_3$ be complex numbers with arguments $2\pi/3$ apart and each has modulus of at most $A>0$. Maximize $|z_1 + z_2 + z_3|$.

Question

Let $a_1 = |z_1|$, $a_2 = |z_2|$, and $a_3 = |z_3|$. Without loss of generality, take

• $z_1 = a_1 $
• $z_2 = a_2 e^{2\pi i/3} $
• $z_3 = a_3 e^{-2\pi i/3}$

We want to maximize $|z_1 + z_2 + z_3|$ subject to

• $|a_1| \leq A$
• $|a_2| \leq A$
• $|a_3| \leq A$

I think the maximum value is $A$. But I don't know how to prove it. Geometrically: we have three complex numbers, 120 degrees apart, each with maximum modulus $A$. If we add them, the result will be inside of the circle centered at origin with radius $A$. Is this correct?

Answer 1

By notation $z_1 = r_1$, $z_2 = r_2e^{2\pi i/3}$ and $z_3 = r_3e^{-2\pi i/3}$, we can write: \begin{align*} |z_1 + z_2 + z_3|^2 &= (z_1 + z_2 + z_3)(\overline{z_1} + \overline{z_2} + \overline{z_3})\\ &= |z_1|^2 + |z_2|^2 + |z_3|^2 + 2\big(\mathrm{Re}(\overline{z_1}z_2) + \mathrm{Re}(\overline{z_2}z_3) + \mathrm{Re}(\overline{z_3}z_1)\big)\\ &= r_1^2 + r_2^2 + r_3^2 + 2\big(r_1r_2\cos(2\pi/3) + r_2r_3\cos(-4\pi/3) + r_3r_1\cos(2\pi/3)\big)\\ &= r_1^2 + r_2^2 + r_3^2 + 2\cos(2\pi/3)\big(r_1r_2 + r_2r_3 + r_3r_1\big)\\ &= r_1^2 + r_2^2 + r_3^2 - r_1r_2 - r_2r_3 - r_3r_1\\ &= \frac{1}{2}\left((r_1 - r_2)^2 + (r_2 - r_3)^2 + (r_3 - r_1)^2\right) \end{align*} Now WLOG assume that $r_1 \leq r_2 \leq r_3$. So there exist $u, v \geq 0$ such that $r_2 = r_1 + u$ and $r_3 = r_1 + u + v$. Therefore: \begin{align*} |z_1 + z_2 + z_3|^2 = \frac{1}{2}\left(u^2 + v^2 + (u + v)^2\right) = u^2 + uv + v^2 \end{align*} Notice that $r_3 \leq A$, so $u + v \leq A$. Consider a triangle with sides $u, v, w$ such that the angle between $u$ and $v$ is fixed and equals $\frac{2\pi}{3}$. By the Law of Cosines, one can easily compute $w^2 = u^2 + uv + v^2$. So $|z_1 + z_2 + z_3| = w$, and we must maximize $w$. Note that if $u + v < A$, then we can increase one of the sides $u$ or $v$, and then $w$ will increase. So we get maximum of $w$ when $u + v = A$. Now for a small number $t$, construct a triangle with sides $u + t$ and $v - t$ in the same direction of $u$ and $v$. Then $(u + t) + (v - t) = A$ holds but now the side $w$ is increased. (See the figure below)

Therefore, if we continue to increase $u$ and decrease $v$, we reach the maximum that is $|z_1 + z_2 + z_3| = w = u + v = A$.

we have $w' > w$.

Answer 2

An edit was necessary due to misunderstanding caused partially by the OP. We have to maximize $|z_{1}+z_{2}+z_{3}|^{2}$

$(a_{1}-\dfrac{1}{2}a_{2}-\dfrac{1}{2}a_{3})^{2}+\dfrac{3}{4}(a_{2}-a_{3})^{2}.$.

Now we notice that the constraint is a cube with sides $[0,A]$. If we try to find an extremum in the interior of the cube, using differentiation, we get a minimum. Therefore the maximum is on the boundary of the cube! Now there are eight corners of the cube. Define $f(z_{1}.z_{2},z_{3})=|z_{1}+z_{2}+z_{3}|$. We take the eight corners of the cube. These are $(0,0,0),(0,0,A),(0,A,0),(A,0,0),(A,A,0),(A,0,A),(0,A,A),(A,A,A).$ All corners give a value $f=A$ except $(0,0,0)$ and $(A,A,A)$.

However, since the function $f$ is convex we have $f(\lambda(z_{1},z_{2},z_{2})+(1-\lambda)(w_{1},w_{2},w_{3}))\leq\,\lambda f(z_{1},z_{2},z_{3})+(1-\lambda)f(w_{1},w_{2},w_{3})$

and using the corners of the cube we can show that points in all segments of the boundary of the cube give a value less or equal than $A$.

Therefore the max is attained on the boundary of the cube and particularly in the corners of the cube (except two) and it is $A$!!

Answer 3

Without loss of generality, assume $a_1\geq a_2$ and $a_1\geq a_3$, and the angle of $z_2$ in relation to $z_1$ is positive. Let $z_1=a_1e^{i\beta}$, so $z_2=a_2e^{i\beta}e^{2\pi i/3}$ and $z_3=a_3e^{i\beta}e^{-2\pi i/3}$.

Now, since $0\leq a_2\leq a_1$, the partial sum $z_1+z_2$ is a point on the line segment between $z_1$ and $z_1e^{2\pi i/6}$:

$$\begin{align}z_1+z_2=\; & a_1e^{i\beta}+a_2e^{i\beta}e^{2\pi i/3} \\ & a_1e^{i\beta}+0e^{i\beta}e^{2\pi i/3}=z_1 \\ & a_1e^{i\beta}+a_1e^{i\beta}e^{2\pi i/3}=z_1(1+e^{2\pi i/3})=z_1e^{2\pi i/6}\end{align}$$

Both of these endpoints are on the circle of radius $a_1$, and thus $z_1+z_2$ is inside that circle.

Likewise, since $0\leq a_3\leq a_1$, the sum $z_1+z_2+z_3$ is a point inside the rhombus with vertices $0$, $z_1$, $z_1e^{2\pi i/6}$, $z_1e^{-2\pi i/6}$:

$$\begin{align}z_1+z_2+z_3=\; & a_1e^{i\beta}+a_2e^{i\beta}e^{2\pi i/3}+a_3e^{i\beta}e^{-2\pi i/3} \\ & a_1e^{i\beta}+0\;e^{i\beta}e^{2\pi i/3}+0\;e^{i\beta}e^{-2\pi i/3}=z_1 \\ & a_1e^{i\beta}+a_1e^{i\beta}e^{2\pi i/3}+0\;e^{i\beta}e^{-2\pi i/3}=z_1(1+e^{2\pi i/3})=z_1e^{2\pi i/6} \\ & a_1e^{i\beta}+0\;e^{i\beta}e^{2\pi i/3}+a_1e^{i\beta}e^{-2\pi i/3}=z_1(1+e^{-2\pi i/3})=z_1e^{-2\pi i/6} \\ & a_1e^{i\beta}+a_1e^{i\beta}e^{2\pi i/3}+a_1e^{i\beta}e^{-2\pi i/3}=z_1(1+e^{2\pi i/3}+e^{-2\pi i/3})=0\end{align}$$

Three of the four vertices are on the cirlce of radius $a_1$, and the fourth vertex is the centre of the circle, so the rhombus is contained in the circle, and thus $z_1+z_2+z_3$ is also.

We see that the maximum value of $|z_1+z_2+z_3|$ is exactly $a_1$, that is, $\max\{|z_1|,|z_2|,|z_3|\}$.

Answer 4

Let $\omega = \exp(2\pi i/3 )$ and $f(a_1, a_2, a_3) = | a_1 + a_2 \omega + a_3 \omega^2|$. Note that $1 + \omega + \omega^2 = 0$, and $f$ is invariant under permutations of the $a_i$, so, wlog, we may assume $a_3 = 0$ and $a_1 > a_2$ and write $f(a, b, 0) = f(a, b)$.

Further, let $\alpha = \exp(\pi i/3 ) = 1 + \omega$, so $f(a,b) = | a - b + b \alpha |$. If we let $b$ range from $0$ to $a$, then $a - b + b \alpha$ takes on the values of the complex numbers on the straight line between $a$ and $a\alpha$, and the moduli of all these numbers are less than $a$, so $f(a,b) \leq \max(a, b)$, $f(a,b, c) \leq \max(a, b, c)$ and the result follows.