 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-1/2c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

w

Re w \leq ∣ w ∣

, while the complex modulus is multiplicative and unchanged by complex conjugation. Consequently

∣ z_{1} + z_{2} ∣^{2} = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} + 2 Re (z_{1} \overline{z_{2}}) \leq ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} + 2∣ z_{1} ∣∣ z_{2} ∣ = (∣ z_{1} ∣ + ∣ z_{2} ∣)^{2} .

Both sides are nonnegative, so taking square roots proves the triangle inequality:

∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣.

The proposed further inequality fails for perpendicular vectors in the complex plane. Taking

z_{1} = 1

z_{2} = i

gives

∣1 + i ∣ + ∣1 - i ∣ = 22 > 2 = 2 max (∣1∣, ∣ i ∣) .

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