The least positive integer $n$ such that $\frac{(2 i)^{n}}{(1-i)^{n-2}}, i=\sqrt{-1}$,\ is a positi

Question

The least positive integer $n$ such that $\frac{(2 i)^{n}}{(1-i)^{n-2}}, i=\sqrt{-1}$,\ is a positive integer, is

Professor PC Thomas Classes · Answer

(6) : Consider $\frac{(2 i)^{n}}{(1-i)^{n-2}}$\ $=2^{n}\left(\frac{i}{1-i}ight)^{n}(1-i)^{2}=2^{n}\left(\frac{i(1+i)}{1-i^{2}}ight)^{n}\left(1+i^{2}-2 iight)$\ $=2^{n} \frac{(-1+i)^{n}}{2^{n}} 	imes(-2 i)=2 	imes(-i)(-1+i)^{n}$\ $=2 	imes(-i) 	imes\left(\sqrt{2} e^{\frac{i 3 \pi}{4}}ight)^{n}=2^{\left(\frac{n}{2}+1ight)} e^{i\left(\frac{3 n \pi}{4}-\frac{\pi}{2}ight)}$\ For this to be positive integer,\ $\frac{3 n \pi}{4}-\frac{\pi}{2}=2 k \pi$, where $k \in I$.\ $\Rightarrow \frac{3 n}{4}=2 k+\frac{1}{2} \Rightarrow n=\frac{2}{3}(4 k+1)$ For $k<0$, we get $n<0$ (not acceptable)\ $k=0$, we get $n=\frac{2}{3} \quad$ (not acceptable)\ $k=1$, we get $n=\frac{10}{3} \quad$ (not acceptable)\ $k=2$, we get $n=\frac{2}{3} 	imes 9=2 	imes 3=6$ (acceptable)\ Hence, the least value of $n$ is 6 .\