Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.9 AMC 10/12 Find the answer China

Complex numbers $z_1, z_2, \dots, z_{100}$ satisfy $z_1 = 3 + 2i$ , $z_{n+1} = \overline{z_n} \cdot i^n$ , ( $n = 1, 2, \dots, 99$ ), with $i$ as the imaginary unit. Then the value of $z_{99} + z_{100}$ is ______.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

By the given conditions, we have
$z_{n+2} = \overline{z_{n+1}} \cdot i^{n+1} = \overline{z_n} \cdot i^n \cdot i^{n+1} = z_n i \quad (n = 1, 2, \dots, 98).$
And since $z_1 = 3 + 2i$ , $z_{99} = z_1 i = -2 + 3i$ . Therefore,

$\begin{aligned} z_{99} + z_{100} &= z_{99} + \overline{z_{99}} \cdot i^{99} \\ &= (-2 + 3i) + (-2 - 3i)(-i) \\ &= -5 + 5i. \end{aligned}$
$\quad \square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.