Animated Solution for Mathematics - Complex Numbers: The least positive integer n for which ${\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)}^n=1$, is :

Beyond the Algebra

My dear students, welcome to a problem that at first glance looks like a tedious algebraic chore. You see a fraction, you see a power $n$, and your instinct might be to reach for the binomial expansion or start rationalizing the denominator with brute force.

But stop. Take a breath. In the world of JEE Advanced, the most elegant solutions are rarely found through brute force; they are found through insight.

Let us look at the expression:

If you try to expand this in rectangular form, you will find yourself drowning in a sea of $i^2$ terms and complex conjugates. It is a path that leads to frustration.

Instead, let us shift our perspective. Let us step into the Argand plane. Imagine you are standing at the origin.

The numerator, $z_1=1+i\sqrt{3}$, is a vector pointing into the first quadrant. Its real part is $1$, and its imaginary part is $\sqrt{3}$.

The denominator, $z_2=1-i\sqrt{3}$, is its mirror image, reflecting perfectly across the real axis into the fourth quadrant.

This is where the magic happens. Whenever you see complex numbers raised to high or unknown powers, Euler's form, $z=r{e}^{i\theta }$, is your best friend. It turns the complex operation of division into a simple subtraction of angles.

Let us calculate the modulus and argument for our numerator, $z_1$. The modulus is $|z_1|=\sqrt{1^2+(\sqrt{3}{)}^2}=2$.

The argument is ${\theta }_1={\tan }^{-1}\left(\frac{\sqrt{3}}{1}\right)=\frac{\pi }{3}$. Thus, $z_1=2e^{i\frac{\pi }{3}}$.

Now, look at the denominator, $z_2$. Because it is a mirror image, the modulus remains $|z_2|=2$, but the argument is simply the negative of the numerator's angle: ${\theta }_2=-\frac{\pi }{3}$.

So, $z_2=2e^{-i\frac{\pi }{3}}$.

Now, watch what happens when we substitute these into our fraction. The ratio becomes:

The modulus $2$ cancels out instantly! We are left with $\frac{e^{i\frac{\pi }{3}}}{e^{-i\frac{\pi }{3}}}$.

Using the laws of exponents, we subtract the denominator's exponent from the numerator's: $i\left(\frac{\pi }{3}-(-\frac{\pi }{3})\right)=i\left(\frac{2\pi }{3}\right)$.

Our complex fraction has simplified to the beautiful, compact form $e^{i\frac{2\pi }{3}}$.

We are now looking for the smallest positive integer $n$ such that $(e^{i\frac{2\pi }{3}}{)}^n=1$. Applying the power rule, we get $e^{i\frac{2n\pi }{3}}=1$.

For a complex exponential to equal $1$, the angle must be a full rotation—or a multiple of a full rotation—around the unit circle. Mathematically, this means the exponent must be an integer multiple of $2\pi i$.

Therefore, we set:

where $k$ is an integer. Dividing both sides by $2\pi$, we find $\frac{n}{3}=k$, or $n=3k$.

To find the least positive integer $n$, we simply choose the smallest positive integer for $k$, which is $k=1$. This gives us $n=3$.

See how the complexity vanished? We didn't need to expand anything. We didn't need to struggle with complex conjugates.

We simply visualized the rotation, applied the power of Euler's form, and let the symmetry of the complex plane guide us to the answer. This is the essence of JEE physics and mathematics—finding the hidden simplicity in the face of apparent complexity.

Keep practicing this mindset, and you will find that no problem is too daunting.