# Find the smallest positive integer n for which

Question:

Find the smallest positive integer $n$ for which $(1+i)^{2 n}=(1-i)^{2 n} .$

Solution:

Given: $(1+i)^{2 n}=(1-i)^{2 n}$

Consider the given equation

$(1+i)^{2 n}=(1-i)^{2 n}$

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

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

Now, rationalizing by multiply and divide by the conjugate of (1 – i)

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

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

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

$\left[(a+b)^{2}=a^{2}+b^{2}+2 a b \&(a-b)(a+b)=\left(a^{2}-b^{2}\right)\right]$

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

$\Rightarrow\left[\frac{2 i}{2}\right]^{2 n}=1$

$\Rightarrow(i)^{2 n}=1$

Now, $\mathrm{i}^{2 \mathrm{n}}=1$ is possible if $\mathrm{n}=2$ because $(\mathrm{i})^{2(2)}=\mathrm{i}^{4}=(-1)^{4}=1$

So, the smallest positive integer n = 2