If ${z^2} + z + 1 = 0$, $z \in C$, then
$$\left| {\sum\limits_{n = 1}^{15} {{{\left( {{z^n} + {{( - 1)}^n}{1 \over {{z^n}}}} \right)}^2}} } \right|$$ is equal to _________.
Answer (integer)
2
Solution
<p>$\because$ ${z^2} + z + 1 = 0$</p>
<p>$\Rightarrow$ $\omega$ or $\omega$<sup>2</sup></p>
<p>$\because$ $$\left| {\sum\limits_{n = 1}^{15} {{{\left( {{z^n} + {{( - 1)}^n}{1 \over {{z^n}}}} \right)}^2}} } \right|$$</p>
<p>$$ = \left| {\sum\limits_{n = 1}^{15} {{z^{2n}} + \sum\limits_{n = 1}^{15} {{z^{ - 2n}} + 2\,.\,\sum\limits_{n = 1}^{15} {{{( - 1)}^n}} } } } \right|$$</p>
<p>$= \left| {0 + 0 - 2} \right|$</p>
<p>$= 2$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
This question is part of PrepWiser's free JEE Main question bank. 223 more solved questions on Complex Numbers and Quadratic Equations are available — start with the harder ones if your accuracy is >70%.