Question

If $|z_1-1|\leq 1$, $|z_2-2|\leq 2$, $|z_3-3|\leq 3$, then find the greatest value of $|z_1+z_2+z_3|$.

• A. 6
• B. 12
• C. 18
• D. 36

Answer 1

Answer: (B)

12

Answer 2

The given inequalities describe closed disks in the complex plane:

1. $|z_1-1|\leq 1$: $z_1$ lies in a disk centered at $c_1=1$ with radius $r_1=1$. The maximum value of $|z_1|$ is $|c_1|+r_1 = |1|+1 = 2$.
2. $|z_2-2|\leq 2$: $z_2$ lies in a disk centered at $c_2=2$ with radius $r_2=2$. The maximum value of $|z_2|$ is $|c_2|+r_2 = |2|+2 = 4$.
3. $|z_3-3|\leq 3$: $z_3$ lies in a disk centered at $c_3=3$ with radius $r_3=3$. The maximum value of $|z_3|$ is $|c_3|+r_3 = |3|+3 = 6$.

By the triangle inequality, $|z_1+z_2+z_3| \leq |z_1|+|z_2|+|z_3|$. The greatest possible value for the sum of the magnitudes is $\max|z_1| + \max|z_2| + \max|z_3| = 2+4+6=12$.

This maximum value is achieved when $z_1$, $z_2$, and $z_3$ are chosen to maximize their individual magnitudes and are aligned in the same direction from the origin. The points that maximize the individual magnitudes are $z_1=2$, $z_2=4$, and $z_3=6$. These points lie on the positive real axis, thus they are collinear and point in the same direction.

Therefore, by choosing $z_1=2$, $z_2=4$, and $z_3=6$, we get $|z_1+z_2+z_3| = |2+4+6| = |12| = 12$.

Alternatively, the set of all possible values for $z_1+z_2+z_3$ is the Minkowski sum of the three disks. The Minkowski sum of disks $D(c_i, r_i)$ is a disk $D(C, R)$ where $C = \sum c_i$ and $R = \sum r_i$. Here, $C = 1+2+3=6$ and $R = 1+2+3=6$. So, $z_1+z_2+z_3$ lies in the disk $|W-6| \leq 6$. The maximum value of $|W|$ for $W$ in this disk is $|C|+R = |6|+6 = 6+6=12$.