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

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to demystify a problem that looks like a tangled mess of fractions but is, in reality, a beautiful, rhythmic dance on the Argand plane.

We are looking for the smallest positive integer $n$ such that:

When you see a complex number in the denominator, your first instinct should be to "clean the house." We cannot easily raise a fraction to a power $n$ if the denominator is complex.

We need to rationalize it. Imagine the denominator $1-i$ as a barrier. To break through it, we multiply both the numerator and the denominator by its complex conjugate, $1+i$.

Why the conjugate? Because $(1-i)(1+i)$ follows the difference of squares identity, $a^2-b^2$. Since $i^2=-1$, this product becomes:

Suddenly, the complex denominator vanishes, leaving us with a simple real number.

Now, look at the numerator: $(1+i)(1+i)$, or $(1+i{)}^2$. Expanding this, we get $1^2+2(1)(i)+i^2$.

Again, we invoke the fundamental law of complex numbers: $i^2=-1$. Substituting this in, our numerator becomes:

When we put it all together, the fraction $\frac{2i}{2}$ simplifies to just $i$. The entire intimidating expression ${\left(\frac{1+i}{1-i}\right)}^n$ has collapsed into the elegant, simple form $i^n$.

Now, we are left with the question: for what smallest positive integer $n$ does $i^n=1$?

Think of $i$ as a rotation operator. Multiplying by $i$ on the Argand plane is equivalent to a $90^{\circ }$ counter-clockwise rotation.

The powers of $i$ follow a cycle: - $i^1=i$ (a $90^{\circ }$ rotation) - $i^2=-1$ (a $180^{\circ }$ rotation) - $i^3=-i$ (a $270^{\circ }$ rotation) - $i^4=1$ (a $360^{\circ }$ rotation, back to the start!)

This cycle repeats every four steps. Therefore, the smallest positive integer $n$ that brings us back to the real value of $1$ is $n=4$.

We look at our options: $n=8$, $n=16$, $n=12$, and "none of these." Since our calculated value of $4$ is not listed, we must confidently select "none of these."

Never fear the "none of these" option. In the JEE, it is not a sign of failure; it is a test of your conviction. You have navigated the algebra, respected the properties of $i$, and arrived at the truth.