Animated Solution for Mathematics - Complex Numbers: Let $z=\frac{1-i\sqrt{3}}{2},i=\sqrt{-1}$. Then the value of $21+{\left(z+\frac{1}{z}\right)}^3+{\left(z^2+\frac{1}{z^2}\right)}^3+{\left(z^3+\frac{1}{z^3}\right)}^3+\cdots +{\left(z^{21}+\frac{1}{z^{21}}\right)}^3$ is.

Welcome, fellow traveler on the road to JEE Advanced. Today, we are not just solving a problem; we are peeling back the layers of a mathematical onion.

When you first look at this expression, it feels like a mountain of algebra. You see $z=\frac{1-i\sqrt{3}}{2}$, and then a massive summation of cubes.

Your instinct might be to panic, to start expanding, or to write pages of calculations. But stop. Breathe. In the world of complex numbers, there is almost always a hidden symmetry waiting to be discovered.

Let us look at $z=\frac{1-i\sqrt{3}}{2}$. If you were to plot this on the Argand plane, you would find the real part is $\frac{1}{2}$ and the imaginary part is $-\frac{\sqrt{3}}{2}$.

This point lies perfectly on the unit circle, meaning its modulus $|z|$ is $1$.

Now, what is the angle? The tangent of the angle is $\frac{-\sqrt{3}/2}{1/2}=-\sqrt{3}$. This corresponds to an angle of $-\frac{\pi }{3}$.

Using Euler's formula, we can write this as $z=e^{-i\frac{\pi }{3}}$. This is our first breakthrough. By converting to exponential form, we have transformed a static algebraic expression into a dynamic, rotating vector.

Look at the general term in our summation: $(z^r+\frac{1}{z^r}{)}^3$. Since $|z|=1$, we know that $\frac{1}{z}=\bar{z}$.

Therefore, $z^r+\frac{1}{z^r}=z^r+{\bar{z}}^r$. Using our exponential form, this becomes $e^{-i\frac{r\pi }{3}}+e^{i\frac{r\pi }{3}}$.

Does this look familiar? It is the classic definition of $2\cos (\theta )$. So, our general term simplifies beautifully to:

We have reduced a complex number expression into a simple trigonometric one. The fear is already starting to dissipate, isn't it?

Now we face $8{\cos }^3(\frac{r\pi }{3})$. We cannot sum cubes of cosines easily, so we must linearize them.

Recall the triple angle identity: $\cos (3\theta )=4{\cos }^3(\theta )-3\cos (\theta )$. Rearranging this, we get:

Multiplying by $2$, we obtain the identity $8{\cos }^3(\theta )=2\cos (3\theta )+6\cos (\theta )$. Substituting $\theta =\frac{r\pi }{3}$, our expression becomes:

Suddenly, the cubic nightmare has vanished, replaced by two simple, linear cosine terms. We are now ready to sum.

The total sum $S$ is $21+{\sum }_{r=1}^{21}(2\cos (r\pi )+6\cos (\frac{r\pi }{3}))$. We can split this into two parts.

First, the alternating series: ${\sum }_{r=1}^{21}2\cos (r\pi )$. As $r$ goes from $1$ to $21$, $\cos (r\pi )$ alternates between $-1$ and $1$.

Since there are $21$ terms (an odd number), the pairs cancel out, leaving us with a single $-1$. Thus, $2\times (-1)=-2$.

Second, the periodic series: ${\sum }_{r=1}^{21}6\cos (\frac{r\pi }{3})$. The cosine function here has a period of $6$.

In any full period of $6$, the sum of these cosine values is $0$. We have $21$ terms, which is $3$ full periods plus $3$ remaining terms ($r=1,2,3$).

For the remaining terms, we calculate:

Multiplying by the coefficient $6$, we get $6\times (-1)=-6$.

Putting it all together, our total sum is $S=21-2-6$.

The result is $13$.

Look at what we have achieved. We started with a terrifying expression involving complex powers and cubes, and through the elegance of Euler's formula, trigonometric identities, and the beauty of periodicity, we arrived at a simple integer.

This is the essence of JEE Advanced mathematics—it is not about brute force; it is about finding the elegant path through the chaos. You have mastered this. Keep that confidence, and carry it into your next challenge.