(A) At $\theta=\frac{\pi}{2}$,we have $\cos \theta = 0$ and $\sin \theta = 1$. Substituting these values into the expression: $\left(\frac{0+i(1)}{1+i

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(A) At $\theta=\frac{\pi}{2}$,we have $\cos \theta = 0$ and $\sin \theta = 1$. Substituting these values into the expression: $\left(\frac{0+i(1)}{1+i(0)}\right)^{2020} + \left(\frac{1+0+i(1)}{1-0+i(1)}\right)^{2021}$ $= (i)^{2020} + \left(\frac{1+i}{1+i}\right)^{2021}$ $= (i^4)^{505} + (1)^{2021}$ $= (1)^{505} + 1 = 1 + 1 = 2$. Since $x+iy = 2$,we have $x=2$ and $y=0$. Therefore,$x+y = 2+0 = 2$.

Class 11 Mathematics 4-1.Complex numbers Mix Examples-Complex numbers

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If $\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{2020}+\left(\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}\right)^{2021} = x+i y$,then the value of $x+y$ at $\theta=\frac{\pi}{2}$ is

TS EAMCET 2020 All TS EAMCET

A
$2$
B
$1$
C
$-1$
D
$2020$

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If $\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{2020}+\left(\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}\right)^{2021} = x+i y$,then the value of $x+y$ at $\theta=\frac{\pi}{2}$ is

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If $\frac{2z_1}{3z_2}$ is a purely imaginary number,then $\left| \frac{z_1 - z_2}{z_1 + z_2} \right| =$

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