Complex Numbers Are Not Really That Complex

$\displaystyle \large \left(\frac{2i}{1+i}\right)^n$

$=\displaystyle \large \left(\frac{1+2i-1}{1+i}\right)^n$

$=\displaystyle \large \left(\frac{1^2 + 2 \cdot 1 \cdot i + i^2 }{1+i}\right)^n$

$=\displaystyle \large \left(\frac{(1+i)^2}{1+i}\right)^n$

$=\displaystyle \large \left(1+i\right)^n$

$=\displaystyle \large \left[\left(1+i\right)^2\right]^{\frac{n}{2}}$

$=\displaystyle \large \left(2i\right)^{\frac{n}{2}}$

$=\displaystyle \large 2^{\frac{n}{2}}\cdot i^{\frac{n}{2}}$

Clearly, for $2^{n/2}$ to be a natural number, its minimum natural power must be $1$ . Therefore $n$ has to be atleast an even number.

Secondly, the minimum power for $i$ to be positive is $4$ . Therefore $\frac{n_{min}}{2}=4 \implies n_{min} = 8$ , which is even and hence satisfies the previous condition.

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Question Source: R. D. Sharma - Mathematics (Class XI) - Complex Numbers - MCQs - Pg. No. 13.64 - Q. No. 13

Relevant wiki: De Moivre's Theorem

$\begin{aligned} f(n) & = \left(\frac {2i}{1+i}\right)^n \\ & = \left(\frac {2i(1-i)}{(1+i)(1-i)}\right)^n \\ & = \left(\frac {2i(1-i)}{2} \right)^n \\ & = \left( i + 1 \right)^n \\ & = \left(\sqrt 2 \left(\frac 1{\sqrt 2} + \frac i{\sqrt 2} \right) \right)^n \\ & = \left(\sqrt 2 \left({\color{#3D99F6}\cos \frac \pi 4 + i \sin \frac \pi 4} \right) \right)^n & \small \color{#3D99F6} \text{By De Moivres theorem} \\ & = 2^\frac n2 \left({\color{#3D99F6} \cos \frac {n \pi}4 + i \sin \frac {n \pi}4} \right) \end{aligned}$

We note that $2^\frac n2$ is a natural number when $n$ is even and $\cos \dfrac {n \pi}4 + i \sin \dfrac {n \pi}4 = 1$ when $\dfrac {n\pi}4 = 2k \pi$ , where $k \in \mathbb N$ , $\implies n = 8k$ , therefore, the smallest $n= \boxed{8}$ .

Use Euler formula: e^(ix) = cos x + i sin x, then〖(2i/(1+i))〗^n = 〖((2e^(i π/2))/(√2e^(i π/4) ))〗^n = 2^(n/2) e^(i*n π/4) . So the minimum value of n such that 〖(2i/(1+i))〗^n is a positive integer is : n=8