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 ${}^n{C}_r$.
$[$ Note $:i^2=-1]$

Answer 1

Answer: 252

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 ${{}^{10}{C}_r]}_{\max .}={}^{10}{C}_5=252$ Ans. $]$