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

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

Express $(-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)$ in the form of $a+ib$.

If $Z_1 = 4i^{40} - 5i^{35} + 6i^{17} + 2$ and $Z_2 = -1 + i$,where $i = \sqrt{-1}$,then $|Z_1 + Z_2| = $

If $x = \frac{5}{1-2i}$,where $i = \sqrt{-1}$,then the value of $x^3 + x^2 - x + 22$ is

If $\sum\limits_{k = 0}^{100} {{i^k}} = x + iy$,then the values of $x$ and $y$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module