Animated Solution for Mathematics - Complex Numbers: The complex number $z=\frac{i-1}{\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}}$ is equal to:

A New Perspective on Complex Numbers

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a journey to understand the hidden geometry of complex numbers.

When you look at the expression $z=\frac{i-1}{\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}}$, your first instinct might be to reach for the conjugate. While that is a valid algebraic path, I want to invite you to see the beauty of the Argand plane.

In the world of complex numbers, division is not just arithmetic—it is a dance of rotation and scaling.

The Numerator as a Vector

Let us focus on the numerator, $z_1=-1+i$. Imagine yourself standing at the origin of the Argand plane. This complex number is a vector pointing to the coordinate $(-1,1)$.

Because the real part is negative and the imaginary part is positive, we are firmly in the second quadrant. To describe this vector in polar form, we need two things: its length (modulus) and its direction (argument).

The modulus is calculated as:

Now, for the angle. The reference angle $\alpha$ is ${\tan }^{-1}|\frac{1}{-1}|=\frac{\pi }{4}$.

Since we are in the second quadrant, our principal argument is ${\theta }_1=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$. Thus, our numerator is:

The Elegance of the Denominator

Now, look at the denominator, $z_2=\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}$. You might recognize this immediately as the polar form of a complex number with modulus $1$ and argument $\frac{\pi }{3}$.

This is the beauty of Euler's form, $e^{i\theta }$. When we see a denominator like this, we should not expand it. We should celebrate it, as it is already prepared for the division operation.

The Magic of Subtraction

Here is where the magic happens. When we divide two complex numbers in polar form, we are performing two distinct operations: we divide their moduli and we subtract their arguments.

Mathematically, if we have the expression $\frac{r_1{e}^{i{\theta }_1}}{r_2{e}^{i{\theta }_2}}$, the result is simply:

Applying this to our problem, the resultant modulus is $r=\frac{\sqrt{2}}{1}=\sqrt{2}$. The resultant argument is $\theta =\frac{3\pi }{4}-\frac{\pi }{3}$.

To subtract these, we find a common denominator of $12$:

The Final Synthesis

By shifting our perspective from tedious algebra to elegant geometry, we have arrived at our answer:

We did not need to expand any brackets or worry about complex conjugates. We simply visualized the vectors, understood their rotation, and performed a subtraction.

This is the mindset of a JEE Advanced topper—always looking for the most elegant, structural path to the solution. Keep practicing this geometric intuition, and you will find that even the most complex problems become simple.