Animated Solution for Mathematics - Complex Numbers: Let $\alpha =\frac{-1+i\sqrt{3}}{2}$. If $a=(1+\alpha ){\sum }_{k=0}^{100}{\alpha }^{2k}$ and $b={\sum }_{k=0}^{100}{\alpha }^{3k}$, then $a$ and $b$ are the roots of the quadratic equation:

Welcome, my fellow math warrior. Today, we are going to dismantle a problem that, at first glance, looks like a terrifying algebraic beast. You see a summation, you see high powers of complex numbers, and your instinct might be to panic.

But I want you to take a deep breath. In JEE Advanced, the most complex-looking problems often hide the most elegant, simple truths. Our goal today is to peel back the layers of this problem and reveal the beauty underneath.

Unmasking the Identity

Look at the value given: $\alpha =\frac{-1+i\sqrt{3}}{2}$. If you have spent time with complex numbers, this should trigger an immediate recognition.

This is not just a random fraction; it is the complex cube root of unity, which we affectionately call $\omega$. Why is this important? Because $\omega$ is a superpower.

It obeys the beautiful cyclic property ${\omega }^3=1$ and the identity $1+\omega +{\omega }^2=0$. By identifying $\alpha =\omega$, we have already won half the battle. We have transformed a scary algebraic expression into a playground of cyclic properties.

The Dance of the Summation

Now, let us look at $a=(1+\alpha ){\sum }_{k=0}^{100}{\alpha }^{2k}$. Substituting $\alpha =\omega$, we get $a=(1+\omega ){\sum }_{k=0}^{100}({\omega }^2{)}^k$.

Do not let the summation sign intimidate you. Let us write out the first few terms: $1+{\omega }^2+{\omega }^4+\cdots +{\omega }^{200}$. This is a classic Geometric Progression (G.P.)!

To sum this, we need three things: the first term $A=1$, the common ratio $R={\omega }^2$, and the number of terms $n$. Here is where students often trip: the index $k$ goes from $0$ to $100$. That means there are $101$ terms.

Using the G.P. sum formula $S=\frac{A(1-R^n)}{1-R}$, we get the sum as:

The Power of Reduction

Now, we face ${\omega }^{202}$. How do we handle such a massive exponent? We use the superpower ${\omega }^3=1$.

We divide $202$ by $3$. Since $202=3\times 67+1$, we know that:

Suddenly, the expression for $a$ becomes:

Look at the numerator: $(1+\omega )(1-\omega )$. This is a difference of squares! It becomes $1-{\omega }^2$.

And look at the denominator: $1-{\omega }^2$. They cancel out perfectly! We are left with $a=1$. Is that not satisfying? The complexity just vanished.

The Simplicity of $b$

Now for $b={\sum }_{k=0}^{100}{\alpha }^{3k}$. Since $\alpha =\omega$, this is ${\sum }_{k=0}^{100}{\omega }^{3k}$.

But wait, ${\omega }^3=1$, so ${\omega }^{3k}=({\omega }^3{)}^k=1^k=1$. The summation is just adding $1$ to itself $101$ times. Thus, $b=101$.

The Final Synthesis

We have our roots: $a=1$ and $b=101$. To form a quadratic equation with roots $a$ and $b$, we use the standard form $x^2-(\text{Sum of roots})x+(\text{Product of roots})=0$.

The sum is $1+101=102$, and the product is $1\times 101=101$. Our final equation is:

See? We started with a daunting expression and ended with a clean, elegant quadratic. This is the essence of JEE Advanced math—it is not about brute force; it is about finding the right perspective. Keep practicing, keep questioning, and keep falling in love with the process.