The least positive integer n such that

Solution:

(b) 8

Let $z=\left(\frac{2 i}{1+i}\right)$

$\Rightarrow z=\frac{2 i}{1+i} \times \frac{1-i}{1-i}$

 

$\Rightarrow z=\frac{2 i(1-i)}{1-i^{2}}$

$\Rightarrow z=\frac{2 i(1-i)}{1+1} \quad\left[\because i^{2}=-1\right]$

$\Rightarrow z=\frac{2 i(1-i)}{2}$

$\Rightarrow z=i-i^{2}$

$\Rightarrow z=i+1$

Now, $z^{n}=(1+i)^{n}$

For $n=2$,

$z^{2}=(1+i)^{2}$

$=1+i^{2}+2 i$

 

$=1-1+2 i$

$=2 i \quad \ldots(1)$

Since this is not a positive integer,

For $n=4$,

$z^{4}=(1+i)^{4}$

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

$=(2 i)^{2}$   [Using (1)]

 

$=4 i^{2}$

$=-4 \quad \ldots(2)$

This is a negative integer.

 

For $n=8$,

$z^{8}=(1+i)^{8}$

$=\left[(1+i)^{4}\right]^{2}$

$=(-4)^{2}$

= 16

This is a positive integer.

Thus, $z=\left(\frac{2 i}{1+i}\right)^{n}$ is positive for $n=8$.

Therefore, 8 is the least positive integer such that $\left(\frac{2 i}{1+i}\right)^{n}$ is a positive integer.