Question

If $z_1$, $z_2$ are two complex numbers such that $|z_{1}| = 1$ and $|z_{2}| = 1$, then the maximum value of $|z_{1} + z_{2}| + |z_{1} - z_{2}|$ is:

• A. 1
• B. 2
• C. $2\sqrt{2}$
• D. 4

Answer 1

Answer: (C)

$2\sqrt{2}$

Answer 2

Let $E = |z_1 + z_2| + |z_1 - z_2|$. We are given that $|z_1| = 1$ and $|z_2| = 1$.

We use the parallelogram law for complex numbers: $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)$.

Substituting the given values, we get $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(1^2 + 1^2) = 2(1 + 1) = 4$.

Let $x = |z_1 + z_2|$ and $y = |z_1 - z_2|$. We know that $x \ge 0$ and $y \ge 0$. The equation becomes $x^2 + y^2 = 4$.

We want to find the maximum value of $E = x + y$. We have $x^2 + y^2 = 4$. We want to maximize $x+y$.

Consider $(x+y)^2 = x^2 + y^2 + 2xy$. Since $x \ge 0$ and $y \ge 0$, $2xy \ge 0$. So $(x+y)^2 = 4 + 2xy$.

To maximize $(x+y)^2$, we need to maximize $xy$. We know that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean: $\frac{x^2+y^2}{2} \ge \sqrt{x^2 y^2} = xy$.

So, $\frac{4}{2} \ge xy$, which means $2 \ge xy$. The maximum value of $xy$ is 2. This occurs when $x^2 = y^2$. Since $x, y \ge 0$, $x=y$.

Substituting $x=y$ into $x^2+y^2=4$, we get $x^2+x^2=4$, so $2x^2=4$, $x^2=2$. Since $x \ge 0$, $x = \sqrt{2}$. Thus $y = \sqrt{2}$.

The maximum value of $xy$ is $\sqrt{2} \times \sqrt{2} = 2$. The maximum value of $(x+y)^2 = 4 + 2xy = 4 + 2(2) = 8$. The maximum value of $x+y$ is $\sqrt{8} = 2\sqrt{2}$.

Alternatively, using Cauchy-Schwarz inequality: $(x+y)^2 = (1 \cdot x + 1 \cdot y)^2 \le (1^2 + 1^2)(x^2 + y^2) = 2(4) = 8$. So, $x+y \le \sqrt{8} = 2\sqrt{2}$. The maximum value is $2\sqrt{2}$.

The equality holds when $\frac{x}{1} = \frac{y}{1}$, i.e., $x=y$. As shown above, $x=y=\sqrt{2}$ satisfies $x^2+y^2=4$.

Therefore, the maximum value of $|z_1 + z_2| + |z_1 - z_2|$ is $2\sqrt{2}$.