Question

If $\omega$ is a non-real imaginary cube root of unity, then find $\arg (i\omega ) + \arg (i{\omega ^2})$

Answer 1

At first we will find the value of $\omega$ and ${\omega ^2}$ using the equation ${x^3} = 1$.
Now, using these values as we substitute the values in the given expression we will simplify for the argument, and simplifying we will get the answer remembering the point that $arguement \in ( - \pi ,\pi ]$ .

Complete step-by-step answer:
Given data: $\arg (i\omega ) + \arg (i{\omega ^2})$ and $\omega$ is a non-real imaginary cube root of unity
It is given that $\omega$ is a non-real imaginary cube root of unity i.e. it is a solution of the equation
${x^3} = 1$
Subtracting 1 from both the sides, we get,
$\Rightarrow {x^3} - 1 = 0$
Using the formula ${a^3} - {b^3} = (a - b)({a^2} + {b^2} + ab)$ , we get,
$\Rightarrow \left( {x - 1} \right)\left( {{x^2} + 1 + x} \right) = 0$
It is given that $\omega$ is the non-real root so $x \ne 1$
$\Rightarrow \left( {{x^2} + 1 + x} \right) = 0$
We know that solution of the quadratic equation $a{x^2} + bx + c = 0$ , is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Therefore, $x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4\left( 1 \right)\left( 1 \right)} }}{2}$
On simplification we get,
$\Rightarrow x = \dfrac{{ - 1}}{2} \pm \dfrac{{\sqrt { - 3} }}{2}$
It is well known that $\sqrt { - 1} = i$
The two values for x will be $\dfrac{{ - 1}}{2} + i\dfrac{{\sqrt 3 }}{2}$ and $\dfrac{{ - 1}}{2} - i\dfrac{{\sqrt 3 }}{2}$
Let us say that, $\omega = \dfrac{{ - 1}}{2} + i\dfrac{{\sqrt 3 }}{2}$ and ${\omega ^2} = \dfrac{{ - 1}}{2} - i\dfrac{{\sqrt 3 }}{2}$
Substituting the value of $\omega$ and ${\omega ^2}$ in $\arg (i\omega ) + \arg (i{\omega ^2})$ , we get,
$= \arg \left[ {i\left( {\dfrac{{ - 1}}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)} \right] + \arg \left[ {i\left( {\dfrac{{ - 1}}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)} \right]$
Simplifying the brackets, we get,
$= \arg \left( {\dfrac{{ - 1}}{2}i + {i^2}\dfrac{{\sqrt 3 }}{2}} \right) + \arg \left( {\dfrac{{ - 1}}{2}i - {i^2}\dfrac{{\sqrt 3 }}{2}} \right)$
Using ${i^2} = - 1$, we get,
$= \arg \left( { - \dfrac{{\sqrt 3 }}{2} - i\dfrac{1}{2}} \right) + \arg \left( {\dfrac{{\sqrt 3 }}{2} - i\dfrac{1}{2}} \right)$
We know that if $\arg \left( {a + ib} \right) = \theta$ then $\tan \theta = \dfrac{b}{a}$ and $- \pi < \arg (z) \leqslant \pi$
$= \arg \left( { - \dfrac{{\sqrt 3 }}{2} - i\dfrac{1}{2}} \right) + \arg \left( {\dfrac{{\sqrt 3 }}{2} - i\dfrac{1}{2}} \right)$
As we know that $\omega$ lie in the third quadrant and ${\omega ^2}$ lie in the second quadrant
$= - \dfrac{{5\pi }}{6} - \dfrac{\pi }{6}$
$= - \pi$, but we mentioned that $arguement \in ( - \pi ,\pi ]$
As point having argument $- \pi$ will lie on –y-axis
The principle argument will be $\pi$
Therefore the required answer is $\pi$ .

Note: We can also do this question with an alternative method
We know that $\arg (a) + \arg (b) = \arg (ab)$ , using this in the simplification of the given expression
$\Rightarrow \arg (i\omega ) + \arg (i{\omega ^2}) = \arg (i\omega i{\omega ^2})$
On simplifying the bracket terms
$= \arg ({i^2}{\omega ^3})$
Now we know that ${i^2} = - 1$ and ${\omega ^3} = 1$ as $\omega$ is a cube root of unity
$= \arg ( - 1)$
Now we know that (-1) lie on the –y-axis whose argument is given by $- \pi$ .