Animated Solution for Mathematics - Complex Numbers: If $z={\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)}^5+{\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)}^5$, then

When you first see the expression $z={\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)}^5+{\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)}^5$, your instinct might be to reach for the binomial expansion. Please, resist that urge! In the JEE Advanced arena, brute force is rarely the intended path.

Let us define our base complex number as $w=\frac{\sqrt{3}}{2}+\frac{i}{2}$. Notice that the second term is its complex conjugate, $\bar{w}=\frac{\sqrt{3}}{2}-\frac{i}{2}$. Our problem is simply $z=w^5+(\bar{w}{)}^5$.

To handle powers like five, we must switch to the language of rotation: Polar form. The modulus of $w$ is:

The argument is $\theta ={\tan }^{-1}\left(\frac{1/2}{\sqrt{3}/2}\right)=\frac{\pi }{6}$. We can now write $w$ in Euler's form as $w=e^{i\frac{\pi }{6}}$.

Because $\bar{w}$ is the conjugate, its angle is simply the negative of the original. Therefore, $\bar{w}=e^{-i\frac{\pi }{6}}$.

Now, apply De Moivre's Theorem. Raising $w$ to the power of five is as simple as multiplying the angle by five:

We are left with the expression $z=e^{i\frac{5\pi }{6}}+e^{-i\frac{5\pi }{6}}$.

Here is the moment of truth. We recall the golden identity:

The imaginary parts, $i\sin \theta$ and $-i\sin \theta$, cancel out perfectly. We are left with:

Since $\frac{5\pi }{6}$ is in the second quadrant, the cosine is negative. Specifically:

Multiplying by two, we get the final result:

The imaginary part is zero. We have navigated the complexity and found a purely real result. This is the power of thinking geometrically.