If left(frac{1-i}{1+i}right)^{100}=a+ib,where | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) First,simplify the expression inside the parenthesis: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+

Vedclass · Accepted Answer

$(1, 0)$

Vedclass · Answer

(A) First,simplify the expression inside the parenthesis: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$. Now,raise this to the power of $100$: $(-i)^{100} = (-1)^{100} 	imes i^{100} = 1 	imes (i^4)^{25} = 1 	imes (1)^{25} = 1$. Given that $a+ib = 1$,we can write this as $1+0i$. Comparing the real and imaginary parts,we get $a=1$ and $b=0$. Thus,$(a, b) = (1, 0)$.

If $\left(\frac{1-i}{1+i}\right)^{100}=a+ib$,where $a, b \in \mathbb{R}$ and $i=\sqrt{-1}$,then $(a, b)$ is equal to

Similar Questions

The value of the series $1 + i^2 + i^4 + i^6 + ..... + i^{2n}$ is:

$i^2+i^3+\ldots+i^{4000}=$

Which of the following statements is true?

The roots of the equation $ix^2 - 4x - 4i = 0$ are

Let $\left(-2-\frac{1}{3} i\right)^3=\frac{x+i y}{27}$,where $i=\sqrt{-1}$ and $x, y$ are real numbers. Then the value of $(y-x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module