Question

Can anyone please solve this​

Answer 1

Topic: Imaginary numbers(Cube roots of unity)

Question:-

What is the value of $\left(\dfrac{-1+\sqrt{3} i}{2} \right)^{900}+\left(\dfrac{-1-\sqrt{3} i}{2} \right)^{301}$?

Hint:-

• Form a quadratic equation having two numbers as a root.
• A quadratic equation with a real coefficient gives complex conjugates as roots. [1]

Solution:-

We observe two numbers are complex conjugates of each other. Let's find which equation gives both numbers as solutions.

Let's try with $\dfrac{-1+\sqrt{3} i}{2}$.

Let $x=\dfrac{-1+\sqrt{3} i}{2}$

$\implies 2x=-1+\sqrt{3} i$

$\implies 2x+1=\sqrt{3} i$

Squaring both sides

$\implies 4x^{2}+4x+1=-3$

Dividing by 4

$\implies \boxed{x^{2}+x+1=0}$

Now, by [1] we know both the numbers are the solutions of the boxed equation since they are complex conjugates.

By multiplying $(x-1)$ we can find that these are the two of the solutions of $x^{3}-1=0$.

$\implies \boxed{x^{3}=1}$

So, both the numbers are the cube roots of unity.

Given:-

$\left\{\left(\dfrac{-1+\sqrt{3} i}{2} \right)^{3}\right\}^{300}+\left\{\left(\dfrac{-1-\sqrt{3} i}{2} \right)^{3}\right\}^{100}\cdot \left(\dfrac{-1-\sqrt{3} i}{2} \right)$

$=(1)^{300}+(1)^{100}\cdot \dfrac{-1-\sqrt{3} i}{2}$

$=1+\dfrac{-1-\sqrt{3} i}{2}$

$=\boxed{\dfrac{1-\sqrt{3} i}{2} }$

This is the required answer.

More information:-

[1] We denote the complex conjugate of $\omega$ as $\bar{\omega}$. To prove this first let's establish an equation.

Given:-

• $ax^{2}+bx+c=0 \ (a,b,c\in \mathbb{R},a\neq 0)$ has $x=\omega,\bar{\omega}$ as solutions.
• $D=b^{2}-4ac<0$

Let's substitute the solution, $\omega$.

$\implies a\omega^{2}+b\omega+c=0$

$\implies \overline{a\omega^{2}+b\omega+c}=0$

$\implies \overline{a\omega^{2}}+\overline{b\omega}+\overline{c}=0$

$\implies a(\overline{\omega})^{2}+b(\overline{\omega})+c=0$

∴ Any real coefficient quadratic equation has a complex number and its conjugate as a pair of solutions if $D<0$.