α,β Roots of x²−x+1=0 — Find α¹⁰¹+β¹⁰⁷

JEE Main 2018. Previous Year Question.

Problem

If $\alpha,\beta\in\mathbb{C}$ are the distinct roots of the equation $x^2-x+1=0$, then $\alpha^{101}+\beta^{107}$ is equal to:

Key insight. The roots of $x^2-x+1=0$ look unfamiliar at first, but comparing them to the well-known cube roots of unity $\omega=\dfrac{-1+\sqrt3 i}{2}$ and $\omega^2=\dfrac{-1-\sqrt3 i}{2}$ reveals that $\alpha$ and $\beta$ are simply $-\omega^2$ and $-\omega$. Once written that way, the huge exponents $101$ and $107$ can be reduced using $\omega^3=1$, turning an intractable power calculation into simple modular arithmetic.

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Approach

Solve the quadratic to get the roots in surd form, then recognise them as $-\omega^2$ and $-\omega$, where $\omega$ is a primitive cube root of unity. Rewrite $\alpha^{101}+\beta^{107}$ using this substitution, reduce each exponent modulo $3$ (since $\omega^3=1$), and use the standard identity $1+\omega+\omega^2=0$ to simplify the result.

Solution

Step 1 — Solve the quadratic

$$x = \frac{1\pm\sqrt{1-4}}{2} = \frac{1\pm\sqrt3\,i}{2}$$

So $\alpha = \dfrac{1+\sqrt3\,i}{2}$ and $\beta = \dfrac{1-\sqrt3\,i}{2}$ (the labelling doesn’t matter since the expression will symmetrically use both).

Step 2 — Recognise these in terms of ω

Recall $\omega = \dfrac{-1+\sqrt3\,i}{2}$ and $\omega^2 = \dfrac{-1-\sqrt3\,i}{2}$. Comparing:

$$\alpha = \frac{1+\sqrt3\,i}{2} = -\left(\frac{-1-\sqrt3\,i}{2}\right) = -\omega^2$$

$$\beta = \frac{1-\sqrt3\,i}{2} = -\left(\frac{-1+\sqrt3\,i}{2}\right) = -\omega$$

Step 3 — Rewrite the target expression

$$\alpha^{101}+\beta^{107} = (-\omega^2)^{101}+(-\omega)^{107} = -\omega^{202}-\omega^{107}$$

(since $(-1)^{101}=-1$ and $(-1)^{107}=-1$)

Step 4 — Reduce the exponents modulo 3

Since $\omega^3=1$:

$$202 = 3\times67+1 \implies \omega^{202} = \omega^1 = \omega$$

$$107 = 3\times35+2 \implies \omega^{107} = \omega^2$$

Step 5 — Substitute back

$$\alpha^{101}+\beta^{107} = -\omega-\omega^2$$

Step 6 — Use the standard identity

Since $1+\omega+\omega^2=0$, we have $\omega+\omega^2=-1$:

$$-\omega-\omega^2 = -(-1) = 1$$

Answer

Common mistakes

• Trying to compute $\alpha^{101}$ directly by repeated multiplication of the surd form. This is essentially impossible by hand; recognising the connection to cube roots of unity is what makes the huge exponents manageable.
• Forgetting the negative sign when writing $\alpha=-\omega^2$, $\beta=-\omega$. Dropping the minus sign changes $(-1)^{101}$ and $(-1)^{107}$ from $-1$ to $+1$, flipping the sign of the final answer.

Practise next

• If $\alpha,\beta$ are the distinct roots of $x^2+x+1=0$, find $\alpha^{100}+\beta^{103}$, using the same cube-roots-of-unity recognition.