The sum of the products of the non-conjugate | vedclass

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

Question

(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 +

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(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 $0$.

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

Similar Questions

If $n$ is an integer and $Z = \cos \theta + i \sin \theta$,where $\theta \neq (2n + 1) \frac{\pi}{2}$,then $\frac{1 + Z^{2n}}{1 - Z^{2n}} = $

If $z = \cos \theta + i \sin \theta$,then $z^r + (\bar{z})^r = $

If $\omega$ is a complex cube root of unity,then $225+(3 \omega+8 \omega^2)^2+(3 \omega^2+8 \omega)^2$ is equal to:

Given $z = (1 + i\sqrt{3})^{100}$,then $\frac{\text{Re}(z)}{\text{Im}(z)}$ equals

If $\alpha$ and $\beta$ are the roots of $x^{2}+x+1=0$,then $\alpha^{16}+\beta^{16}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module