Animated Solution for Mathematics - Complex Numbers: Let $\alpha =\frac{-1+i\sqrt{3}}{2}$ is 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 :

Imagine you are standing at the threshold of a complex problem. You see $\alpha =\frac{-1+i\sqrt{3}}{2}$ and your first instinct might be to panic, to start plugging in values, or to try some brute-force algebra.

But stop. In the world of JEE Advanced, this specific number is a beacon. It is $\omega$, the complex cube root of unity.

Recognizing this is your first step toward mastery. It transforms a nightmare of calculation into a dance of elegant properties. We know that:

These are not just equations; they are the keys to the kingdom.

Let us look at $b={\sum }_{k=0}^{100}{\alpha }^{3k}$. Since $\alpha =\omega$, this becomes ${\sum }_{k=0}^{100}({\omega }^3{)}^k$.

Because ${\omega }^3=1$, this simplifies to ${\sum }_{k=0}^{100}{1}^k$. We are simply adding $1$ to itself, over and over.

From $k=0$ to $k=100$, there are exactly $101$ terms. So, $b=101$. The complexity vanishes when you apply the right identity.

Now, we turn our attention to $a=(1+\alpha ){\sum }_{k=0}^{100}{\alpha }^{2k}$. Substituting $\alpha =\omega$, we get $a=(1+\omega ){\sum }_{k=0}^{100}{\omega }^{2k}$.

The summation is a Geometric Progression where the first term $A=1$, the common ratio $r={\omega }^2$, and the number of terms $n=101$. Using the sum formula $S=\frac{A(r^n-1)}{r-1}$, we get:

Now, we use our reduction rule. Since $202=3\times 67+1$, we have ${\omega }^{202}=\omega$.

Thus, the sum becomes:

Factoring the denominator as $(\omega -1)(\omega +1)$, we see the $(\omega -1)$ terms cancel out, leaving $S=\frac{1}{\omega +1}$.

Finally, $a=(1+\omega )\cdot \frac{1}{\omega +1}$, which simplifies to $a=1$.

We have found our roots: $a=1$ and $b=101$. To form the quadratic equation, 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 $1\cdot 101=101$.

Our final equation is:

By identifying the core identity early, we turned a terrifying problem into a series of logical, satisfying steps. Keep this mindset, and no problem will ever be too daunting.