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  4. If ( ( 1 - i/1 + i))96 = a + ib then (a,b) is

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Q. If $\left( \frac{ 1 - i}{1 + i}\right)^{96} = a + ib $ then (a,b) is

KCETKCET 2018Complex Numbers and Quadratic Equations

Solution:

$\frac{1-i}{1+i} = \frac{1-i}{1+i} \times\frac{1-i}{1-i} = \frac{1+i^{2} -2i}{1-i^{2}} =-i$
$ \therefore \left(\frac{1-i}{1+i}\right)^{96} = a +ib$
$ \Rightarrow \left(-i\right)^{96} =a + ib$
$ \Rightarrow 1 =a + ib$
$\therefore \left(a,b\right)\equiv \left(1,0\right)$