Animated Solution for Mathematics - Complex Numbers: If ${\left(\frac{1+i}{1-i}\right)}^{m/2}={\left(\frac{1+i}{i-1}\right)}^{n/3}=1$, $(m,n\in N)$ then the greatest common divisor of the least values of $m$ and $n$ is

My dear student, welcome to this journey through the elegant world of complex numbers. When you first look at an equation like

it is natural to feel a slight shiver of intimidation. Those fractional powers and complex bases look like a storm of variables.

But let us take a deep breath. In the JEE Advanced, the most complex-looking problems often hide the most beautiful, simple truths. Our goal is not to fight the equation, but to peel back its layers.

Cleansing the Denominator

We begin by focusing on the base expressions. We have $\frac{1+i}{1-i}$. To simplify this, we use the classic technique of rationalization by multiplying the numerator and the denominator by the complex conjugate of the denominator, which is $1+i$.

This gives us:

Expanding the numerator, we get $1^2+2i+i^2$. Since $i^2=-1$, this simplifies to $2i$, and the denominator becomes $1-(-1)=2$.

Thus, the entire fraction collapses into $i$. It is a moment of pure satisfaction when that bulky fraction simplifies into a single, elegant unit on the imaginary axis.

Now, look at the second base: $\frac{1+i}{i-1}$. If you look closely, you will see that $i-1$ is simply $-(1-i)$.

Therefore, the fraction is $-\left(\frac{1+i}{1-i}\right)$. Since we already know $\frac{1+i}{1-i}=i$, the second base is simply $-i$. We have bypassed the heavy lifting with a single observation.

The Cyclic Heartbeat of $i$

Now, our equation is transformed into something much friendlier: $i^{m/2}=1$ and $(-i{)}^{n/3}=1$. This brings us to the heartbeat of complex numbers: the cyclic nature of powers.

We know that $i^1=i$, $i^2=-1$, $i^3=-i$, and $i^4=1$. This cycle repeats every four powers. For $i^k=1$, $k$ must be a multiple of 4.

Similarly, for $(-i{)}^k=1$, we see that $(-i{)}^1=-i$, $(-i{)}^2=-1$, $(-i{)}^3=i$, and $(-i{)}^4=1$. So, for both bases, the exponent must be a multiple of 4.

Solving for $m$ and $n$

We are looking for the least natural values of $m$ and $n$. For the first condition, $\frac{m}{2}$ must be the smallest positive multiple of 4, which is 4.

Thus, $\frac{m}{2}=4$, which gives $m=8$. For the second condition, $\frac{n}{3}$ must also be 4, giving $n=12$.

We have found our values!

The Final GCD

Finally, the problem asks for the greatest common divisor of $m=8$ and $n=12$. The factors of 8 are 1, 2, 4, and 8.

The factors of 12 are 1, 2, 3, 4, 6, and 12. The largest common factor is 4.

And there you have it—the answer is 4. You see, my friend, the complexity was just a mask. With patience and the right tools, even the most daunting problems yield to the beauty of logic.