Question

Let ${\left(\frac{1-i}{1+i}\right)}^{100}=a+ib$, then the value of $a,b$ are respectively,

• A. $0,1$
• B. $1,0$
• C. $-1.0$
• D. $0,-1$

Answer 1

Answer: (B)

$1,0$

We have,

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

$\Rightarrow {\left(\frac{(1-i{)}^2}{(1+i{)}^2}\right)}^{50}=a+ib$

$\Rightarrow {\left(\frac{-2i}{2i}\right)}^{50}=a+ib$ [ Since $(1+i{)}^2=2i$ and $(1-i{)}^2=-2i$]

$1=a+ib$

Comparing both sides we get,

$a=1,b=0$.