Animated Solution for Mathematics - Complex Numbers: If $z=\frac{\sqrt{3}}{2}+\frac{i}{2}$, then $(1+iz+z^5+i{z}^8{)}^9$ is equal to

When you first look at the expression $(1+iz+z^5+i{z}^8{)}^9$, it might seem like a chaotic mess of powers and imaginary units. In the world of JEE Advanced, chaos is often just order in disguise. Let's peel back the layers.

Our journey begins with $z=\frac{\sqrt{3}}{2}+\frac{i}{2}$. The first thing a master student does is look for the geometry. Calculate the magnitude:

This is our first clue! Our complex number $z$ lives on the unit circle. This means we can represent $z$ as $e^{i\theta }$.

Since $\cos \theta =\frac{\sqrt{3}}{2}$ and $\sin \theta =\frac{1}{2}$, we know $\theta =\frac{\pi }{6}$. Thus, $z=e^{i\pi /6}$. This is the key that unlocks the entire problem.

Now, let's look at the terms inside the bracket: $1$, $iz$, $z^5$, and $i{z}^8$. Instead of struggling with Cartesian powers, we use Euler's identity.

Multiplying by $i$ is just a $90^{\circ }$ counter-clockwise turn. Similarly, $z^5=(e^{i\pi /6}{)}^5=e^{i5\pi /6}$.

Finally, $i{z}^8=e^{i\pi /2}\cdot (e^{i\pi /6}{)}^8=e^{i\pi /2}\cdot e^{i4\pi /3}=e^{i11\pi /6}$. See how the exponents just add up?

Now, let's convert these back to Cartesian form to see what happens when we sum them. Let $S=1+iz+z^5+i{z}^8$.

We have $iz=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$, $z^5=-\frac{\sqrt{3}}{2}+\frac{i}{2}$, and $i{z}^8=\frac{\sqrt{3}}{2}-\frac{i}{2}$.

Look closely at $z^5$ and $i{z}^8$. They are exact opposites! When you add them, they vanish into thin air: $z^5+i{z}^8=0$.

The scary expression collapses into:

And what is $\frac{1}{2}+i\frac{\sqrt{3}}{2}$? It is simply $e^{i\pi /3}$.

We are left with $(e^{i\pi /3}{)}^9$. Using the laws of exponents, this is:

We know that $e^{i3\pi }=\cos (3\pi )+i\sin (3\pi )$. Since $\cos (3\pi )=-1$ and $\sin (3\pi )=0$, our final answer is -1.