Question

If $|z_1|=1, |z_2|=2, |z_3|=3$ and $|9z_1\overline{z_2}+4z_1\overline{z_3}+z_2\overline{z_3}|=12$, then find the value of $|z_1+z_2+z_3|$.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

Let the given complex numbers be $z_1, z_2, z_3$. We are given their magnitudes: $|z_1|=1$ $|z_2|=2$ $|z_3|=3$

From these, we have $z_1\overline{z_1} = |z_1|^2 = 1$, $z_2\overline{z_2} = |z_2|^2 = 4$, and $z_3\overline{z_3} = |z_3|^2 = 9$. This implies $\overline{z_1} = 1/z_1$, $\overline{z_2} = 4/z_2$, and $\overline{z_3} = 9/z_3$.

We are also given the condition $|9z_1\overline{z_2}+4z_1\overline{z_3}+z_2\overline{z_3}|=12$. Let $u = 9z_1\overline{z_2}$, $v = 4z_1\overline{z_3}$, and $w = z_2\overline{z_3}$. Let's find the magnitudes of these terms: $|u| = |9z_1\overline{z_2}| = 9|z_1||\overline{z_2}| = 9 \times 1 \times 2 = 18$. $|v| = |4z_1\overline{z_3}| = 4|z_1||\overline{z_3}| = 4 \times 1 \times 3 = 12$. $|w| = |z_2\overline{z_3}| = |z_2||\overline{z_3}| = 2 \times 3 = 6$.

The given condition is $|u+v+w|=12$. We have magnitudes $|u|=18, |v|=12, |w|=6$. For the sum of three complex numbers, $|u+v+w| \le |u|+|v|+|w|$. $12 \le 18+12+6 = 36$. This is true.

Consider the case where the complex numbers $u, v, w$ are collinear. The value 12 can be obtained if $u$ and $w$ are in the same direction, and $v$ is in the opposite direction, and their magnitudes add up as $18+6-12=12$. This implies that $u$ and $w$ are in the same direction, and $v$ is in the opposite direction. So, $\arg(u) = \arg(w)$ and $\arg(v) = \arg(u) + \pi$.

From $\arg(u) = \arg(w)$, we have $u = k w$ for some $k>0$. $|u| = k|w| \implies 18 = k \times 6 \implies k=3$. So, $u = 3w$. $9z_1\overline{z_2} = 3(z_2\overline{z_3})$ $3z_1\overline{z_2} = z_2\overline{z_3}$.

From $\arg(v) = \arg(u) + \pi$, we have $v = -m u$ for some $m>0$. $|v| = m|u| \implies 12 = m \times 18 \implies m=12/18 = 2/3$. So, $v = -\frac{2}{3}u$. $4z_1\overline{z_3} = -\frac{2}{3}(9z_1\overline{z_2})$ $4z_1\overline{z_3} = -6z_1\overline{z_2}$ Since $z_1 \neq 0$, we can divide by $z_1$: $4\overline{z_3} = -6\overline{z_2}$. $\overline{z_3} = -\frac{3}{2}\overline{z_2}$. Taking conjugates, $z_3 = -\frac{3}{2}z_2$.

Now we have two relations:

1. $3z_1\overline{z_2} = z_2\overline{z_3}$
2. $z_3 = -\frac{3}{2}z_2$

Substitute (2) into (1): $3z_1\overline{z_2} = z_2\overline{(-\frac{3}{2}z_2)}$ $3z_1\overline{z_2} = z_2(-\frac{3}{2}\overline{z_2})$ $3z_1\overline{z_2} = -\frac{3}{2}z_2\overline{z_2}$ $3z_1\overline{z_2} = -\frac{3}{2}|z_2|^2$ $3z_1\overline{z_2} = -\frac{3}{2}(4)$ $3z_1\overline{z_2} = -6$ $z_1\overline{z_2} = -2$.

Now we want to find $|z_1+z_2+z_3|$. Let $S = z_1+z_2+z_3$. $|S|^2 = (z_1+z_2+z_3)(\overline{z_1}+\overline{z_2}+\overline{z_3})$ $|S|^2 = |z_1|^2+|z_2|^2+|z_3|^2 + (z_1\overline{z_2}+z_2\overline{z_1}) + (z_1\overline{z_3}+z_3\overline{z_1}) + (z_2\overline{z_3}+z_3\overline{z_2})$ $|S|^2 = |z_1|^2+|z_2|^2+|z_3|^2 + 2\text{Re}(z_1\overline{z_2}) + 2\text{Re}(z_1\overline{z_3}) + 2\text{Re}(z_2\overline{z_3})$.

We found $z_1\overline{z_2} = -2$. So $\text{Re}(z_1\overline{z_2}) = -2$. From $z_3 = -\frac{3}{2}z_2$, we have $\overline{z_3} = -\frac{3}{2}\overline{z_2}$. $z_1\overline{z_3} = z_1(-\frac{3}{2}\overline{z_2}) = -\frac{3}{2}(z_1\overline{z_2}) = -\frac{3}{2}(-2) = 3$. So $\text{Re}(z_1\overline{z_3}) = 3$. $z_2\overline{z_3} = z_2(-\frac{3}{2}\overline{z_2}) = -\frac{3}{2}z_2\overline{z_2} = -\frac{3}{2}|z_2|^2 = -\frac{3}{2}(4) = -6$. So $\text{Re}(z_2\overline{z_3}) = -6$.

Substitute these values into the expression for $|S|^2$: $|S|^2 = 1+4+9 + 2(-2) + 2(3) + 2(-6)$ $|S|^2 = 14 - 4 + 6 - 12$ $|S|^2 = 10 + 6 - 12$ $|S|^2 = 16 - 12$ $|S|^2 = 4$.

Therefore, $|S| = |z_1+z_2+z_3| = \sqrt{4} = 2$.