Complex numbers $z$, $w$, $u$ with $|z|=1$, $|w|=2$, $|u|=3$, Find the maximum value of $|z-u|$

Question

Let be given three complex numbers $z$, $w$, $u$ satisfying $|z|=1$, $|w|=2$, $|u|=3$, and $|z+w-u|=|u+z-w|$. Find the greatest value of $|z-u|$.

I tried $$|z+w-u| \leqslant |z-u| + |w| \leqslant |z + u| + |w| \leqslant |z| + |u| + |w|.$$ This way doesn't mention the condition $|z+w-u|=|u+z-w|$. How can I solve the prolblem?

Answer 1

Let $z=e^{i\alpha}, w=2e^{i\beta}, u=3e^{i\gamma}$, we are given $|z+w-u|=|u+z-w|$, re-write as

$$|w-u-(-z)|=|w-u-z|$$

This means the number $w-u$ is located on the perpendicular bisector line of $z$ and $-z$, hence

$$\frac{\Im(w-u)}{\Re(w-u)}=-\frac{1}{\tan\alpha}$$

Simplify and we get

$$3\cos(\alpha-\gamma)=2\cos(\alpha-\beta)\tag{1}$$

Next, we want to find the maximum of $|z-u|$

$$|z-u|=\sqrt{(\cos\alpha-3\cos\gamma)^2+(\sin\alpha-3\sin\gamma)^2}=\sqrt{10-6\cos(\alpha-\gamma)}$$

Plug in (1)

$$|z-u|=\sqrt{(\cos\alpha-3\cos\gamma)^2+(\sin\alpha-3\sin\gamma)^2}=\sqrt{10-4\cos(\alpha-\beta)}$$

We get the maximum

$$\max|z-u|=\sqrt{14}$$

Answer 2

Another way, as a slight variation to the fine solution already given by Anne Bauval, assume wlog $z=1$ then by $v=w-u$

$$|z+w-u|=|u+z-w| \iff |1+v|=|1-v|$$

that is $v=iy$, then assuming wlog $w=x+(y-a)i$ and $u=-x+ia$ with $x\ge 0$ we have

• $|u|^2=x^2+a^2=9 \implies x=\sqrt{9-a^2}$
• $|w|^2=x^2+y^2-2ay+a^2=4\implies y=a\pm\sqrt{a^2-5}$

from which we find that $5\le a^2\le 9$ and since $z-u=1+x-ia$ we have

$$|z-u|^2=(1+x)^2+a^2=1+x^2+2x+a^2=10+2\sqrt{9-a^2}$$

which is maximized for $a^2=5$ giving $|z-u|=\sqrt{14}$.

Answer 3

Alternative using geometry on Argand plane, define two points:

$$ \begin{aligned} P_{1} &= z+u-w \\ P_{2} &= z+w-u \end{aligned} $$

The plot on complex plane looks like the following:

To satisfy requirement $|P_{1}|=|P_{2}|$, the vector $\vec{z}_{1}$ must be perpendicular to $\vec{w_{1}}-\vec{u_{1}}$. Then to maximize the objective, we simply put $u_{1}$ (and consequently, $w_{1}$ as far away from $z_{1}$ as possible as shown in the following:

The rest is simply elementary geometry.

Answer 4

Wlog $z=i.$

$$|i+w-u|=|i+u-w|\iff a:=w-u\in\Bbb R.$$

Set $u=x+iy$ (with $x,y\in\Bbb R$). Since $x^2+y^2=9$ and $(x+a)^2+y^2=4,$ $$x=\frac{-a^2-5}{2a}.$$ The minimum value of $|x|$ is therefore $\sqrt5$ (for $a=\pm\sqrt5$) and the minimum value of $y$ is $-\sqrt{9-5}=-2,$ hence the maximum value of $|i-u|=\sqrt{10-2y}$ is $$\sqrt{10+2\cdot2}=\sqrt{14}.$$ Note that the corresponding value of $w$ is $$w=-2z,$$ which suggest a geometric solution.

Answer 5

I have an idea based on the answer of @MathFail, but more intuitive.

Let $z=e^{i\alpha}$, $w=2e^{i\beta}$, $u=3e^{i\gamma}$.

In this question, we only care about the modulus of $z$, $w$, $u$, and addition/subtraction between them. As all constraints are about the modulus, only the relative value of arguments $\alpha-\beta$, $\alpha-\gamma$ have an effect. So we can let $\alpha=0$, i.e. $z=1$, without any loss of generality.

Considering this problem on the complex plane, as is discussed in the answer of @MathFail, $u-w$ is perpendicular to $z$, which means $\Re(u-w)=\Re(w+(-u))=0$.

Given the $|z-u|=|z+(-u)|$ reaches its maximum value when $\gamma=\pi$ and decreases along with the decreasing (increasing) of $\gamma$ from $\pi$ to 0 ($2\pi$)

However, $\Re(u-w)=\Re(w+(-u))=0$ limits the possible maximum values of $\gamma$. The maximum $abs(\Re(-u))$ to cancel out the $\Re(w)$ is $2$. In this condition, the $\max(\gamma)$ is $\arccos(-\frac{4}{9})$ and $\max|z-u|=\sqrt{14}$