If (x+iy) = left(frac{1+i}{1-i}right)^3 - lef | vedclass

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

Question

(B) Given $(x+iy) = \left(\frac{1+i}{1-i}\right)^3 - \left(\frac{1-i}{1+i}\right)^3$. First,simplify the base fractions: $\frac{1+i}{1-i} = \frac{(1+i

Vedclass · Accepted Answer

$x > y$

Vedclass · Answer

(B) Given $(x+iy) = \left(\frac{1+i}{1-i}\right)^3 - \left(\frac{1-i}{1+i}\right)^3$. First,simplify the base fractions: $\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{1+2i-1}{1+1} = \frac{2i}{2} = i$. Similarly,$\frac{1-i}{1+i} = \frac{1}{i} = -i$. Substituting these into the expression: $x+iy = (i)^3 - (-i)^3$. $x+iy = -i - (-i^3) = -i - (i) = -2i$. Comparing the real and imaginary parts: $x = 0$ and $y = -2$. Since $0 > -2$,we have $x > y$.

If $(x+iy) = \left(\frac{1+i}{1-i}\right)^3 - \left(\frac{1-i}{1+i}\right)^3$,then the true statement among the following is

Similar Questions

Express the given complex number in the form $a+ib$: $i^{9}+i^{19}$

The simplified form of $i^{n} + i^{n+1} + i^{n+2} + i^{n+3}$ is

The least positive integer $n$ such that $\frac{(2i)^{n}}{(1-i)^{n-2}}$,where $i=\sqrt{-1}$,is a positive integer,is ..... .

Express $(5-3i)^{3}$ in the form $a+ib$.

The values of $\theta$,for which $\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is real are

Vedclass Test Series

Exam Paper Generator

Online Exam Module