Question:
If $\left( \frac{ 1 - i}{1 + i}\right)^{96} = a + ib $ then (a,b) is
If $\left( \frac{ 1 - i}{1 + i}\right)^{96} = a + ib $ then (a,b) is
Updated On: Aug 15, 2024
- (1,1)
- (1,0)
- (0,1)
- (0,-1)
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The Correct Option is B
Solution and Explanation
$\frac{1-i}{1+i} = \frac{1-i}{1+i} \times\frac{1-i}{1-i} = \frac{1+i^{2} -2i}{1-i^{2}} =-i$
$ \therefore \left(\frac{1-i}{1+i}\right)^{96} = a +ib$
$ \Rightarrow \left(-i\right)^{96} =a + ib$
$ \Rightarrow 1 =a + ib$
$\therefore \left(a,b\right)\equiv \left(1,0\right)$
$ \therefore \left(\frac{1-i}{1+i}\right)^{96} = a +ib$
$ \Rightarrow \left(-i\right)^{96} =a + ib$
$ \Rightarrow 1 =a + ib$
$\therefore \left(a,b\right)\equiv \left(1,0\right)$
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
