If (√3 + i)¹⁰⁰ = 2⁹⁹(p + iq), then p and q are roots of the equation : (1) x² - (√3 - 1)x - √3 = 0 (

Q: Wait, why can't I just use the Binomial Theorem? It's a standard method, right?

The Binomial Theorem is a powerful tool, but it is not the *right* tool here. Expanding a binomial to the power of 100 would result in 101 terms, most of which would be complex and messy. In JEE Advanced, efficiency is as important as accuracy. Euler's form turns a massive algebraic expansion into a simple multiplication problem.

Q: How do I know if I should use $2\pi$ or $4\pi$ when reducing the angle?

Great question! You are looking for the principal argument. You want to subtract the largest multiple of $2\pi$ that is less than your current angle. Since $16\pi$ is a multiple of $2\pi$, it represents 8 full rotations. Removing it brings you back to the same position on the Argand plane without changing the value of the complex number.

Q: Why did we factor out $2^{99}$ at the end?

We did that to match the structure of the given equation: $2^{99}(p + iq)$. By forcing our result to look like the target expression, we make the comparison between the real and imaginary parts trivial. Always look at the 'target' form in the question—it's a roadmap for your final steps.

Question

If $(\sqrt{3} + i)^{100} = 2^{99}(p + iq)$, then $p$ and $q$ are roots of the equation :

Options:

(A) $x^2 - (\sqrt{3} - 1)x - \sqrt{3} = 0$

(B) $x^2 + (\sqrt{3} + 1)x + \sqrt{3} = 0$

(C) $x^2 + (\sqrt{3} - 1)x - \sqrt{3} = 0$

(D) $x^2 - (\sqrt{3} + 1)x + \sqrt{3} = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare. You see $(\sqrt{3} + i)^{100}$ and your instinct might be to panic.

If you feel the urge to expand this binomial a hundred times, take a deep breath. In the world of JEE Advanced, we don't use brute force; we use elegance. We use the geometry of the complex plane.

### Phase 1: The Argand Plane Visualization

Let us define our complex number as $z = \sqrt{3} + i$. Before we touch any algebra, let's look at the geometry. Imagine the Argand plane where we move $\sqrt{3}$ units along the real axis and $1$ unit up the imaginary axis.

To find the modulus $r$, we use the Pythagorean theorem: 
$$r = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$$

Now, for the argument $	heta$. We know that $	an(	heta) = \frac{1}{\sqrt{3}}$, which implies $	heta = \frac{\pi}{6}$. We have successfully translated our Cartesian coordinate into a polar coordinate: $(r, 	heta) = (2, \frac{\pi}{6})$.

### Phase 2: The Euler Transformation

Now, we invoke one of the most beautiful identities in mathematics: Euler's Formula, $z = re^{i	heta}$. Instead of dealing with the messy $(\sqrt{3} + i)$, we write our number as $z = 2e^{i\frac{\pi}{6}}$.

When we raise this to the power of $100$, we use the laws of exponents rather than binomial expansion:

$$z^{100} = (2e^{i\frac{\pi}{6}})^{100} = 2^{100} \cdot (e^{i\frac{\pi}{6}})^{100}$$

The power of $100$ distributes to the $2$, and for the exponential term, we use the rule $(e^{a})^b = e^{ab}$. This is De Moivre's Theorem in its most natural habitat.

### Phase 3: The Reduction of the Angle

We are left with $2^{100} \cdot e^{i\frac{100\pi}{6}}$. Simplifying the exponent, we get $\frac{50\pi}{3}$.

Since adding $2\pi$ brings you back to the same spot in the complex plane, we find the principal angle by writing:

$$\frac{50\pi}{3} = \frac{48\pi + 2\pi}{3} = 16\pi + \frac{2\pi}{3}$$

Because $16\pi$ represents $8$ full rotations, it effectively vanishes. Our expression simplifies to $2^{100} e^{i\frac{2\pi}{3}}$.

### Phase 4: Returning to Cartesian Reality

We convert back from Euler's form to Cartesian form using $e^{i	heta} = \cos(	heta) + i\sin(	heta)$:

$$2^{100} \left( \cos\left(\frac{2\pi}{3}ight) + i\sin\left(\frac{2\pi}{3}ight) ight)$$

Substituting the trigonometric values $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$ and $\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$, we get:

$$2^{100} \left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} ight) = 2^{99}(-1 + i\sqrt{3})$$

Comparing this to $2^{99}(p + iq)$, we identify the constants as **$p = -1$** and **$q = \sqrt{3}$**.

### Phase 5: The Final Quadratic Construction

We construct a quadratic equation $x^2 - Sx + P = 0$ using the sum $S = p + q$ and product $P = pq$.

The sum is $S = -1 + \sqrt{3} = \sqrt{3} - 1$. The product is $P = (-1)(\sqrt{3}) = -\sqrt{3}$.

Plugging these into our standard form, we obtain the final equation:

$$x^2 - (\sqrt{3} - 1)x - \sqrt{3} = 0$$

We didn't fight the problem; we flowed with it. By using geometry to simplify the algebra, we bypassed the tedious calculations to reach the **final quadratic equation**.

Animated Solution for Mathematics - Complex Numbers: If $(3 + i)^{100} = 2^{99} (p + i q)$ , then $p$ and $q$ are roots of the equation :

Select Answer:

Visualized Solution

Visualizing $z = 3 + i$

The Power Problem

Calculating Modulus

Calculating Argument

Euler Form Setup

Raising to Power $100$

Simplifying the Exponent

Reducing the Angle

Back to Cartesian Form

Evaluating Trig Values

Matching the RHS

Extracting $p$ and $q$

Sum and Product of Roots

The Final Quadratic Equation

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

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (3​+i)100=299(p+iq), then p and q are roots of the equation :