frac{(1+i)^{2011}}{(1-i)^{2009}} is equal to | vedclass

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

Question

(D) We know that $(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$. Given expression: $E = \frac{(1+i)^{2011}}{(1-i)^{2009}} = \frac{(1

Vedclass · Accepted Answer

-$2$

Vedclass · Answer

(D) We know that $(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$. Given expression: $E = \frac{(1+i)^{2011}}{(1-i)^{2009}} = \frac{(1+i)^{2009} \cdot (1+i)^2}{(1-i)^{2009}}$. $E = \left(\frac{1+i}{1-i}\right)^{2009} \cdot (1+i)^2$. Simplify $\frac{1+i}{1-i}$: $\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{1+i^2+2i}{1-i^2} = \frac{2i}{2} = i$. Substitute back: $E = (i)^{2009} \cdot (2i) = 2 \cdot i^{2010}$. Since $i^4 = 1$,$i^{2010} = (i^4)^{502} \cdot i^2 = 1^{502} \cdot (-1) = -1$. Therefore,$E = 2 \cdot (-1) = -2$.

$\frac{(1+i)^{2011}}{(1-i)^{2009}}$ is equal to

Similar Questions

If $z$ and $z'$ are complex numbers such that $z \cdot z' = z$,then $z' = $

The value of the sum $\sum\limits_{n = 1}^{13} {({i^n} + {i^{n + 1}})} $,where $i = \sqrt { - 1} $,equals

Express the following in the form $a+ib$:
$i^{-35}$

Express the given complex number in the form $a+ib$: $(1-i)-(-1+i6)$

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