If n is a positive integer and frac{(1+i)^n}{ | vedclass

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(B) Given,$\frac{(1+i)^n}{(1-i)^n} = -i$ $\Rightarrow \left(\frac{1+i}{1-i}\right)^n = -i$ Rationalizing the denominator inside the bracket: $\Rightar

Vedclass · Accepted Answer

$4k-1, k \in N$

Vedclass · Answer

(B) Given,$\frac{(1+i)^n}{(1-i)^n} = -i$ $\Rightarrow \left(\frac{1+i}{1-i}\right)^n = -i$ Rationalizing the denominator inside the bracket: $\Rightarrow \left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^n = -i$ $\Rightarrow \left[\frac{1+i^2+2i}{1-i^2}\right]^n = -i$ Since $i^2 = -1$: $\Rightarrow \left[\frac{1-1+2i}{1+1}\right]^n = -i$ $\Rightarrow \left(\frac{2i}{2}\right)^n = -i$ $\Rightarrow i^n = -i$ We know that $i^1 = i$,$i^2 = -1$,$i^3 = -i$,and $i^4 = 1$. The value $i^n = -i$ occurs when $n$ is of the form $4k-1$ for $k \in N$ (e.g.,for $k=1, n=3$; for $k=2, n=7$).

If $n$ is a positive integer and $\frac{(1+i)^n}{(1-i)^n} = -i$,then $n$ will be of the form:

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