A complex number z among the following which | vedclass

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

Question

(C) Given,$z^3+27 i=0$. Since $27 i = (-3 i)^3$,we have $z^3 - (-3 i)^3 = 0$. Using the identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$,we get $(z - (-3 i))(

Vedclass · Accepted Answer

$(3 \sqrt{3}+3 i) / 2$

Vedclass · Answer

(C) Given,$z^3+27 i=0$. Since $27 i = (-3 i)^3$,we have $z^3 - (-3 i)^3 = 0$. Using the identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$,we get $(z - (-3 i))(z^2 + z(-3 i) + (-3 i)^2) = 0$. $(z + 3 i)(z^2 - 3 i z - 9) = 0$. Case $1$: $z + 3 i = 0 \Rightarrow z = -3 i$. Case $2$: $z^2 - 3 i z - 9 = 0$. Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $z = \frac{3 i \pm \sqrt{(-3 i)^2 - 4(1)(-9)}}{2} = \frac{3 i \pm \sqrt{-9 + 36}}{2} = \frac{3 i \pm \sqrt{27}}{2} = \frac{3 i \pm 3 \sqrt{3}}{2}$. Thus,the roots are $-3 i$,$\frac{3 \sqrt{3} + 3 i}{2}$,and $\frac{-3 \sqrt{3} + 3 i}{2}$. Comparing with the options,the correct value is $\frac{3 \sqrt{3} + 3 i}{2}$.

$A$ complex number $z$ among the following which satisfies $z^3+27 i=0$ is

Similar Questions

If $\omega$ is a complex cube root of unity and $x = \omega^2 - \omega + 2$,then:

If $z(2-i)=(3+i)$,then $z^{38} = ?$ (where $z=x+iy$)

If $i{z^4} + 1 = 0$,then $z$ can take the value

If ${z_1}, {z_2}, {z_3}, ......, {z_n}$ are the $n^{th}$ roots of unity,then for $k = 1, 2, ....., n-1$:

If $\sqrt{x}+\frac{1}{\sqrt{x}}=2 \cos \theta$,then $x^6+x^{-6}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module