Question

w²+w³+w⁴ if w is a complex cube root of unity, find the value​

Answer 1

Answer:

Hope it helps you a lot.......

Answer 2

Given data: $w^{2}+w^{3}+w^{4}$

To Find: The value, if w is a complex cube root of unity.

Solution:

Cube roots of unity contains three roots $1,w,w^{2}$ and the value of $w^{3}=1$.

According to cube roots of unity, $1+w+w^{2}=0$, where $1+w+w^{2}$ are the sum of the cube roots of unity.

Consider the given cube root of unity,

$=w^{2}+w^{3}+w^{4}$

Expand $w^{4}$ as cube root of unity $w^{3}.w$

$=w^{2}+w^{3}+(w^{3}.w)$

Substitute, $w^{3}=1$,

$=w^{2}+1+(1.w)$

$=w^{2}+1+w$

Rearrange the terms from lowest power to highest power,

$=1+w+w^{2}$

$=0$, since $1+w+w^{2}=0$.

Therefore, if w is the complex cube root of unity, then the value of $w^{2}+w^{3}+w^{4}$ is $0$.