(B) Let $z = i^{1/4}$,which implies $z^4 = i = e^{i\pi/2}$.The equation is $z^4 - i = 0$.The roots of this equation are $z_k = e^{i\left(\frac{\pi/2 +

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(B) Let $z = i^{1/4}$,which implies $z^4 = i = e^{i\pi/2}$.The equation is $z^4 - i = 0$.The roots of this equation are $z_k = e^{i\left(\frac{\pi/2 + 2k\pi}{4}\right)}$ for $k = 0, 1, 2, 3$.These roots are $e^{i\pi/8}, e^{i5\pi/8}, e^{i9\pi/8}, e^{i13\pi/8}$.None of these roots are conjugates of each other.According to Vieta's formulas for the polynomial $z^4 + 0z^3 + 0z^2 + 0z - i = 0$,the sum of the products of the roots taken two at a time is the coefficient of $z^2$,which is $0$.Therefore,the sum is $0$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers De Moivre's theorem and Roots of unity

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The sum of the products of the non-conjugate roots of $i^{1/4}$ taken two at a time is

TS EAMCET 2019 All TS EAMCET

A
$2$
B
$0$
C
$-1$
D
$-2$

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Class 11

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Chapter

1 Complex Numbers and Quadratic Equations: Complex Numbers

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De Moivre's theorem and Roots of unity

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If ${\left( {\frac{{1 + \cos \theta + i\sin \theta }}{{i + \sin \theta + i\cos \theta }}} \right)^4} = \cos n\theta + i\sin n\theta $,then $n$ is equal to

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The number of complex roots of the equation $x^{11}-x^7+x^4-1=0$ whose arguments lie in the first quadrant is

EasyTS EAMCET 2018

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If $\alpha$ is a non-real root of the equation $x^6-1=0$,then $\frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1} = $

DifficultTS EAMCET 2012

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If $\alpha, \beta$ are the roots of the equation $x+\frac{4}{x}=2 \sqrt{3}$,then $\frac{2}{\sqrt{3}}\left|\alpha^{2024}-\beta^{2024}\right|=$

MediumTS EAMCET 2024

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Let $z_k = \cos \left(\frac{2k\pi}{10}\right) + i \sin \left(\frac{2k\pi}{10}\right); k = 1, 2, \ldots, 9$.

List-$I$ List-$II$

$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$ $1.$ True

$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers. $2.$ False

$R.$ $\frac{|1-z_1||1-z_2| \ldots |1-z_9|}{10}$ equals $3.$ $1$

$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals $4.$ $2$

Codes: $P \quad Q \quad R \quad S$

IIT 2014

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List-$I$	List-$II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$	$1.$ True
$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\|1-z_1\|\|1-z_2\| \ldots \|1-z_9\|}{10}$ equals	$3.$ $1$
$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals	$4.$ $2$

The sum of the products of the non-conjugate roots of $i^{1/4}$ taken two at a time is

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