Animated Solution for Mathematics - Complex Numbers: If $i=\sqrt{-1}$, then $4+5{\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)}^{334}+3{\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)}^{365}$ is equal to

Welcome, fellow traveler on the JEE journey. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of high-power arithmetic.

You see an expression like $4+5{\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)}^{334}+3{\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)}^{365}$ and your instinct might be to panic. But here is the secret of the JEE Advanced: the problem setter is not testing your ability to calculate massive powers; they are testing your ability to recognize patterns.

Recognizing $\omega$

Let us look at the term inside the parentheses: $z=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$. If you have spent time in the complex plane, this should feel like meeting an old friend. This is the primitive cube root of unity, which we denote as $\omega$.

Imagine standing at the origin of the complex plane. If you draw a unit circle, this number $\omega$ sits exactly at an angle of $120^{\circ }$ (or $\frac{2\pi }{3}$ radians) from the positive real axis.

It is a rotation operator. When you multiply a complex number by $\omega$, you are essentially rotating it by $120^{\circ }$. This is why ${\omega }^3=1$—if you rotate by $120^{\circ }$ three times, you have completed a full $360^{\circ }$ circle and returned to the starting point, $1$. This is the heartbeat of the problem.

We are faced with exponents $334$ and $365$. If we try to calculate these directly, we will be here until the next JEE cycle. Instead, we use the property ${\omega }^3=1$.

This means that any power of $\omega$ can be reduced by simply looking at the remainder when the exponent is divided by $3$. For the first term, we have ${\omega }^{334}$. We divide $334$ by $3$:

This tells us that ${\omega }^{334}=({\omega }^3{)}^{111}\cdot {\omega }^1=1^{111}\cdot \omega =\omega$.

For the second term, we have ${\omega }^{365}$. We divide $365$ by $3$:

This tells us that ${\omega }^{365}=({\omega }^3{)}^{121}\cdot {\omega }^2=1^{121}\cdot {\omega }^2={\omega }^2$. Just like that, the terrifying exponents have vanished, leaving us with a simple quadratic-like expression: $4+5\omega +3{\omega }^2$.

Now, we have $4+5\omega +3{\omega }^2$. We know another fundamental property of the cube roots of unity: the sum of all three roots is zero. That is, $1+\omega +{\omega }^2=0$.

We want to use this identity to simplify our expression. Let us break down the coefficients to create groups of $(1+\omega +{\omega }^2)$. We can rewrite the expression as follows:

Look closely at the terms $3+3\omega +3{\omega }^2$. We can factor out the $3$:

Since $1+\omega +{\omega }^2=0$, the entire term $3(1+\omega +{\omega }^2)$ becomes zero! We are left with the elegant, simple result: $1+2\omega$.

We are almost there. Now, we simply substitute the value of $\omega$ back into our simplified expression:

Distributing the $2$ gives us:

The $1$ and $-1$ cancel out perfectly, leaving us with the final answer: $i\sqrt{3}$.

Look at what we have achieved. We started with a complex, intimidating expression and, through the lens of symmetry and periodicity, reduced it to a single imaginary term.

This is the essence of JEE Advanced physics and mathematics. It is never about the brute force; it is about finding the underlying structure. You have successfully navigated the complex plane, utilized the cyclic properties of roots, and arrived at the solution with grace. Keep this mindset—always look for the pattern, always trust the identity, and you will conquer any problem they throw at you.