Roots of x²+2x+2=0 — Find α¹⁵+β¹⁵

JEE Main 2019 — 9 January, Morning Shift. Previous Year Question.

Problem

Let $\alpha$ and $\beta$ be two roots of the equation $x^2+2x+2=0$. Then $\alpha^{15}+\beta^{15}$ is equal to:

Key insight. Raising a complex number to the 15th power directly (in $a+bi$ form) would be a nightmare of repeated multiplication. Converting to polar form first turns “raise to the 15th power” into “multiply the angle by 15 and raise the modulus to the 15th power” — De Moivre’s theorem — which is vastly simpler.

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Approach

Solve the quadratic to find $\alpha$ and $\beta$ in the form $a+bi$, then convert each to polar form (modulus and argument). Apply De Moivre’s theorem to raise each to the 15th power, reduce the resulting angle to its simplest equivalent, and add the two results (which will be complex conjugates of each other, so their sum is purely real).

Solution

Step 1 — Solve the quadratic

$$x = \frac{-2\pm\sqrt{4-8}}{2} = \frac{-2\pm2i}{2} = -1\pm i$$

So $\alpha=-1+i$ and $\beta=-1-i$.

Step 2 — Convert to polar form

For $\alpha=-1+i$: modulus $=\sqrt{(-1)^2+1^2}=\sqrt2$, and since it lies in the second quadrant, argument $=\dfrac{3\pi}{4}$.

$$\alpha = \sqrt2\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)$$

Since $\beta$ is the complex conjugate of $\alpha$:

$$\beta = \sqrt2\left(\cos\frac{3\pi}{4}-i\sin\frac{3\pi}{4}\right)$$

Step 3 — Apply De Moivre’s theorem to α¹⁵

$$\alpha^{15} = (\sqrt2)^{15}\left(\cos\frac{45\pi}{4}+i\sin\frac{45\pi}{4}\right)$$

Step 4 — Reduce the angle modulo 2π

$$\frac{45\pi}{4} \div 2\pi = \frac{45}{8} = 5.625$$

Subtracting $5$ full rotations ($5\times2\pi=\frac{40\pi}{4}$):

$$\frac{45\pi}{4}-\frac{40\pi}{4} = \frac{5\pi}{4}$$

Step 5 — Simplify α¹⁵

$$(\sqrt2)^{15} = 2^{7.5} = 128\sqrt2$$

$$\alpha^{15} = 128\sqrt2\left(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\right) = 128\sqrt2\left(-\frac{\sqrt2}{2}-i\frac{\sqrt2}{2}\right) = -128-128i$$

Step 6 — Find β¹⁵ using the conjugate relationship

Since $\beta$ is the conjugate of $\alpha$, $\beta^{15}$ is the conjugate of $\alpha^{15}$:

$$\beta^{15} = -128+128i$$

Step 7 — Add the two results

$$\alpha^{15}+\beta^{15} = (-128-128i)+(-128+128i) = -256$$

Answer

Common mistakes

• Trying to multiply out $(-1+i)^{15}$ by repeated binomial expansion. This is technically possible but extremely error-prone; converting to polar form first is far more reliable for high powers.
• Forgetting to reduce the angle modulo $2\pi$ before evaluating sine and cosine. Leaving the angle as $\dfrac{45\pi}{4}$ and trying to evaluate trig functions directly is unnecessarily error-prone compared to first reducing it to an angle within one full rotation.

Practise next

• Let $\gamma,\delta$ be the roots of $x^2-2x+2=0$. Find $\gamma^{10}+\delta^{10}$, using the same polar-form and De Moivre’s theorem approach.