The least positive integer n such that frac{( | vedclass

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(C) Given expression: $E = \frac{(2i)^{n}}{(1-i)^{n-2}}$First,simplify $(1-i)^2 = 1 + i^2 - 2i = 1 - 1 - 2i = -2i$.Thus,$(1-i)^{n-2} = ((1-i)^2)^{\fra

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Given expression: $E = \frac{(2i)^{n}}{(1-i)^{n-2}}$First,simplify $(1-i)^2 = 1 + i^2 - 2i = 1 - 1 - 2i = -2i$.Thus,$(1-i)^{n-2} = ((1-i)^2)^{\frac{n-2}{2}} = (-2i)^{\frac{n-2}{2}}$.Substituting this into the expression: $E = \frac{(2i)^n}{(-2i)^{\frac{n-2}{2}}} = \frac{(2i)^n}{(-1)^{\frac{n-2}{2}} (2i)^{\frac{n-2}{2}}} = \frac{(2i)^{n - \frac{n-2}{2}}}{(-1)^{\frac{n-2}{2}}} = \frac{(2i)^{\frac{n+2}{2}}}{(-1)^{\frac{n-2}{2}}}$.For $E$ to be a positive integer,the power of $i$ must be a multiple of $4$ (i.e.,$\frac{n+2}{2} = 4k$) and the magnitude must be real.If $n=6$,$\frac{n+2}{2} = 4$. Then $E = \frac{(2i)^4}{(-1)^2} = \frac{16 i^4}{1} = 16(1) = 16$,which is a positive integer.Thus,the least positive integer $n$ is $6$.

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

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