Prove the finite complex-sum identity for positive even n under the stated angle restrictions.

01 / Standard#Your turn

w=\cos\theta+i\sin\theta,\quad0<\theta<2\pi,\ \theta\ne\pi

Official paper · jm02-2025 · 4(c) · PDF 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Apply the finite geometric-sum formula.

Hint 2

Factor e^{it}−1 as 2ie^{it/2}sin(t/2).

Worked solution

Since w≠1, the sum has a nonzero denominator.
$\sum_{k=0}^nw^k=\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}$
Factor numerator and denominator and simplify the exponential phase.
$\frac{2ie^{i(n+1)\theta/2}\sin((n+1)\theta/2)}{2ie^{i\theta/2}\sin(\theta/2)}=e^{in\theta/2}\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}=w^{n/2}\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}$

The required identity holds; n/2 is an integer and sin(θ/2)≠0.

Checks and common pitfalls: Using exponential phases avoids ambiguity from choosing complex square roots of w.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Since w≠1, the sum has a nonzero denominator.
$\sum_{k=0}^nw^k=\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}$
Factor numerator and denominator and simplify the exponential phase.
$\frac{2ie^{i(n+1)\theta/2}\sin((n+1)\theta/2)}{2ie^{i\theta/2}\sin(\theta/2)}=e^{in\theta/2}\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}=w^{n/2}\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}$

Think first. Reveal a hint when the class is ready.

2025 JM02