Question

If ${\left(\frac{1-i}{1+i}\right)}^{100}=a+ib$, then find $(a,b)$.

Answer 1

Given :
${\left(\frac{1-i}{1+i}\right)}^{100}=a+ib$

$L.H.S.={\left(\frac{1-i}{1+i}\right)}^{100}$
$\Rightarrow L.H.S.={(\frac{(1-i)}{(1+i)}\times \frac{(1-i)}{(1-i)})}^{100}$
$\Rightarrow L.H.S.={\left(\frac{1+i^2-2i}{1-i^2}\right)}^{100}$
$\Rightarrow L.H.S.={\left(\frac{-2i}{2}\right)}^{100}[\because i^2=-1]$
$\Rightarrow L.H.S.=(-i{)}^{100}=(i{)}^{100}\times (-1{)}^{100}$
$\Rightarrow L.H.S.=i^{100}$
$\Rightarrow L.H.S.=(i^4{)}^{25}$
$\Rightarrow L.H.S.=1\{\because i^4=1\}$

Comparing $L.H.S.$ and $R.H.S.$
$\because L.H.S.=R.H.S.$
$1=a+ib$

Comparing corresponding real and imaginary part
$\Rightarrow a=1,b=0$

Hence, the value of $(a,b)$ is $(1,0)$