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Question: Let $z_1, z_2 \in C$ be complex numbers such that $|z_1 + z_2| = \sqrt{3}$ and $|z_1| = |z_2| = 1$. ...

Let z1,z2∈Cz_1, z_2 \in C be complex numbers such that ∣z1+z2∣=3|z_1 + z_2| = \sqrt{3} and ∣z1∣=∣z2∣=1|z_1| = |z_2| = 1. Compute ∣z1−z2∣|z_1 - z_2|.

Answer

1

Explanation

Solution

The problem can be solved using the parallelogram law for complex numbers, which states that for any two complex numbers z1z_1 and z2z_2: ∣z1+z2∣2+∣z1−z2∣2=2(∣z1∣2+∣z2∣2)|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2) Given the conditions: ∣z1∣=1|z_1| = 1 ∣z2∣=1|z_2| = 1 ∣z1+z2∣=3|z_1 + z_2| = \sqrt{3}

Substitute these values into the parallelogram law: (3)2+∣z1−z2∣2=2(12+12)(\sqrt{3})^2 + |z_1 - z_2|^2 = 2(1^2 + 1^2) 3+∣z1−z2∣2=2(1+1)3 + |z_1 - z_2|^2 = 2(1 + 1) 3+∣z1−z2∣2=2(2)3 + |z_1 - z_2|^2 = 2(2) 3+∣z1−z2∣2=43 + |z_1 - z_2|^2 = 4 Subtracting 3 from both sides: ∣z1−z2∣2=4−3|z_1 - z_2|^2 = 4 - 3 ∣z1−z2∣2=1|z_1 - z_2|^2 = 1 Taking the square root of both sides (and noting that modulus is always non-negative): ∣z1−z2∣=1|z_1 - z_2| = 1