Let α, β be the roots of the equation x² - √6x + 3 = 0 such that Im(α) Im(β). Let a, b be integers

Q: Wait, why do we reject the negative root for the imaginary part? How do we know which is $\alpha$ and which is $\beta$?

The problem explicitly gives the condition $\text{Im}(\alpha) > \text{Im}(\beta)$. Since the roots are complex conjugates, one will have a positive imaginary part and the other a negative one. By assigning the positive angle $\frac{\pi}{4}$ to $\alpha$, we satisfy the condition.

Q: Is it really necessary to use Euler's form? Can't I just expand $\alpha^{99}$ using the binomial theorem?

Expanding $\alpha^{99}$ using the binomial theorem would be a nightmare of calculations! In JEE Advanced, whenever you see high powers of complex numbers, Euler's form ($e^{i\theta}$) is the intended path. It turns exponentiation into simple multiplication of the angle.

Q: Why does the expression simplify so cleanly? Is there a hidden trick?

It's not a trick, but the beauty of symmetry! By factoring out $\alpha^{98}$, we reveal the sum of the roots, $\alpha + \beta$, which is a coefficient from the original quadratic. This is a classic pattern in JEE problems designed to reward those who look for algebraic structure before diving into raw calculation.

Question

Let $\alpha, \beta$ be the roots of the equation $x^2 - \sqrt{6}x + 3 = 0$ such that $	ext{Im}(\alpha) > 	ext{Im}(\beta)$. Let $a, b$ be integers not divisible by 3 and $n$ be a natural number such that $\frac{\alpha^{99}}{\beta} + \alpha^{98} = 3^n(a + ib), i = \sqrt{-1}$. Then $n + a + b$ is equal to

Sigma Percentile · Accepted Answer

### The Beauty of Complex Symmetry

Welcome, future engineers! Today, we are tackling a problem that might look like a brute-force calculation nightmare, but is actually a masterclass in elegance.

We are given the quadratic equation $x^2 - \sqrt{6}x + 3 = 0$ and asked to evaluate a massive expression involving the 99th power of its roots. If you try to calculate $\alpha^{99}$ directly, you will be lost in a sea of arithmetic. Let's find the smarter way.

### Phase 1: The Quadratic Foundation

First, let's find our roots. Using the quadratic formula

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

we substitute $a=1$, $b=-\sqrt{6}$, and $c=3$.

The discriminant is $D = (-\sqrt{6})^2 - 4(1)(3) = 6 - 12 = -6$. This confirms our roots are complex:

$$x = \frac{\sqrt{6} \pm i\sqrt{6}}{2}$$

By factoring out $\sqrt{6}$, we get $x = \sqrt{6} \left( \frac{1 \pm i}{2} ight)$. With a little manipulation, this becomes $x = \sqrt{3} \left( \frac{1 \pm i}{\sqrt{2}} ight)$. This is the moment where your intuition should scream: Euler's form!

### Phase 2: The Euler Transformation

We recognize $\frac{1 \pm i}{\sqrt{2}}$ as $\cos(\frac{\pi}{4}) \pm i\sin(\frac{\pi}{4})$, which is $e^{\pm i\pi/4}$. Thus, our roots are $\alpha = \sqrt{3}e^{i\pi/4}$ and $\beta = \sqrt{3}e^{-i\pi/4}$.

The condition $	ext{Im}(\alpha) > 	ext{Im}(\beta)$ forces $\alpha$ to be the root with the positive angle. Now, look at the target expression:

$$E = \frac{\alpha^{99}}{\beta} + \alpha^{98}$$

Instead of calculating powers, let's factor:

$$E = \alpha^{98} \left( \frac{\alpha}{\beta} + 1 ight) = \alpha^{98} \left( \frac{\alpha + \beta}{\beta} ight)$$

### Phase 3: The Elegant Cancellation

We know from Vieta's formulas that $\alpha + \beta = \sqrt{6}$. Substituting this into our expression, we get

$$E = \frac{\alpha^{98} \cdot \sqrt{6}}{\beta}$$

Now, substitute the Euler forms:

$$E = \frac{(\sqrt{3}e^{i\pi/4})^{98} \cdot \sqrt{6}}{\sqrt{3}e^{-i\pi/4}}$$

The magnitude part is $(\sqrt{3})^{98} \cdot \frac{\sqrt{6}}{\sqrt{3}} = 3^{49} \cdot \sqrt{2}$. The exponential part is

$$\frac{e^{i 98\pi/4}}{e^{-i\pi/4}} = e^{i(98\pi/4 + \pi/4)} = e^{i 99\pi/4}$$

Since $\frac{99\pi}{4} = 24\pi + \frac{3\pi}{4}$, and $24\pi$ is just 12 full rotations, we are left with $e^{i 3\pi/4}$.

### Phase 4: The Final Victory

We have

$$E = 3^{49} \cdot \sqrt{2} \cdot \left(\cos\left(\frac{3\pi}{4}ight) + i\sin\left(\frac{3\pi}{4}ight)ight)$$

Since $\cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}}$ and $\sin(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}}$, the expression becomes

$$E = 3^{49} \cdot \sqrt{2} \cdot \left(-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}ight) = 3^{49}(-1 + i)$$

Comparing this to $3^n(a + ib)$, we find $n=49$, $a=-1$, and $b=1$. The final sum $n+a+b = 49 - 1 + 1 = \mathbf{49}$. See? No brute force, just pure, beautiful logic.

Enter Numerical Value:

Visualized Solution

$A na l y z e t h e Q u a d r a t i c E q u a t i o n$

$T h e Q u a d r a t i c F or m u l a$

$E v a l u a t in g t h eD i scr iminan t$

$S im pl i f y in g t h e R oo t s$

$C o n v er tt o E u l e r^{'} s F or m$

$I d e n t i f y α an d β$

$S im pl i f y t h e T a r g e tE x p r ess i o n$

$S u b s t i t u t e S u m o f R oo t s$

$S u b s t i t u t e E u l er F or m s$

$S im pl i f y M a g ni t u d es$

$C o mbin e E x p o n e n t ia l T er m s$

$E v a l u a t e t h e T r i g o n o m e t r i c P a r t$

$C o m p a r e an d S o l v e$

The Sigma Insight: Euler's Form and De Moivre's Theorem

The Beauty of Complex Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AnalyzetheQuadraticEquation

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ConverttoEuler′sForm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifyαandβ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifytheTargetExpression

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SubstituteEulerForms

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyMagnitudes

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CombineExponentialTerms

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CompareandSolve