Question

Write the two complex cube root of 1​

Answer 1

Cube Roots Of Unity. Let the cube root of 1 be x i.e., 3√ 1 = x. Hence, there are three cube roots of unity which are which the first one is real and the othertwo are conjugate complex numbers. These complex cube roots of unity are also called imaginary cube roots of unity.

Answer 2

Concept:

If the two  solutions are complex, then the formulas for all three solutions contain the cube roots of real  real numbers, but if all three solutions are real, they can be represented by the complex roots of  complex numbers.

Given

Given number is $1.$

Find

We need to find  two complex cube roots in the $1.$

Solution

The cube root of unity is equivalent to a variable$z.$

$\sqrt[3]{1}=z$

For example, the cube and the cube root of a number are inverse operations. Therefore, if you move the cube root  to the opposite side of the equation, you get a cube with the opposite number.

$1=z^{3}$

Shift $1$  also moves to the other side of the equation. Therefore, the value of LHS is zero.

$z^{3}-1=0$

From the algebraic identity element $a^3-b^3 = (a - b) (a^2+ ab + b^2)$, of factor $z^3-1$.

$(z-1) (z^2 + z + 1)=0$        ......(1)

Simplify the coefficients further to calculate the value of   $z$.

From the equation (1) either $z-1=0$ or $z^2+ z + 1= 0$.

If $z-1=0$ then $z=1$

$z^2+ z + 1= 0$ is simplified using a formula for solving quadratic equations.

According to equation,

$Z=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

in the above equation, the general form of the quadratic equation is considered to be  $ax^2+bx+c=0$. Comparing the general equations with and $z^2 + z + 1= 0, a = 1, b = 1,$and $c=1.$

assign these values ​​to Equation

$\begin{aligned}Z&=\frac{-1\pm \sqrt{1^2-4\times 1\times 1}}{2\times 1}\\ Z&=\frac{-1\pm \sqrt{-3}}{2}\end$

Therefore, the complex cube roots of $1$ obtained by solving  $z^2 + z + 1= 0$are  $-\frac{1}{2}-\frac{\sqrt{3}}{2}$ and $-\frac{1}{2}+\frac{\sqrt{3}}{2}$

However, $\sqrt{-1}=i$

substitutes the roots obtained above the three values ​​of the cube.

The roots of unity is: $\sqrt[3]{1}=1,-\frac{1}{2}+i\frac{\sqrt{3}}{2},-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

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