Question

Let $z =i \omega^2$ where $\omega$ is the imaginary cube root of unity and $n$ is the smallest positive integer for which $\left( z ^{89}+i^{97}\right)^{94}= z ^{ n }$ then find the maximum value of ${ }^{ n } C _{ r }$.
$\left[\right.$ Note $\left.: i^2=-1\right]$

Answer 1

complexiOMR We have $z =i \omega^2$
Given L.H.S. $=\left( z ^{89}+i^{97}\right)^{94}=\left(\left(i \omega^2\right)^{89}+i\right)^{94}=\left(i \omega^{178}+i\right)^{94}=(i \omega+i)^{94}=i^{94}\left(-\omega^2\right)^{94}=-\omega^{188}=-\omega^2$
Given R.H.S $= z ^{ n }=\left(i \omega^2\right)^{ n }$ hence $\left(i \omega^2\right)^{ n }=-\omega^2$
$\Rightarrow $ Least positive integral value of $n$ is 10 .
Hence $\left.{ }^{10} C _{ r }\right]_{\max .}={ }^{10} C _5=252$ Ans. $]$