Question

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

• A. $a=2,b=-1$
• B. $a=1,b=0$
• C. $a=0,b=1$
• D. $a=1,b=2$

Answer 1

Answer: (B)

$a=1,b=0$

${\left(\frac{1-i}{1+i}\right)}^{100}$$=(-i{)}^{100}$ ($\because \left(\frac{1-i}{1+i}\right)=-i$ )
$\Rightarrow 1=a+ib$ ($\because i^{4n}=1$ )
$\therefore a=1$ and $b=0$
Hence, option B.