Animated Solution for Mathematics - Complex Numbers: If $z=\frac{\sqrt{3}}{2}+\frac{i}{2},i=\sqrt{-1}$, then $(z^{201}-i{)}^8$ is equal to

Imagine you are standing on the complex plane, looking at the number $z=\frac{\sqrt{3}}{2}+\frac{i}{2}$. In the context of JEE Advanced, this is a gateway to a beautiful geometric rotation.

When you see a massive exponent like $201$, the secret to conquering this problem lies not in brute-force algebra, but in the elegant language of Euler's formula.

Before we touch the exponent, we must understand the nature of $z$. We calculate its magnitude:

Our complex number sits perfectly on the unit circle. Now, we find its argument, $\theta$:

We have successfully identified $z$ as a unit vector pointing at $30^{\circ }$.

Now, we invoke our secret weapon: Euler's formula, $z=r{e}^{i\theta }$. With $r=1$ and $\theta =\frac{\pi }{6}$, our complex number transforms into the compact form:

This is the key that unlocks the entire problem. Instead of dealing with real and imaginary parts, we are now dealing with a simple rotation.

We need to calculate $z^{201}$. Using our Euler form, this becomes:

Simplifying the fraction $\frac{201}{6}$ gives $\frac{67}{2}$, or $33.5$. Thus, our expression is $e^{i(33.5\pi )}$, which we can split as:

We know that $e^{i(33\pi )}=-1$ because $33$ is an odd multiple of $\pi$. Since $e^{i\frac{\pi }{2}}=i$, we find:

We are now ready to evaluate the original expression $(z^{201}-i{)}^8$. Substituting our result, we get:

We expand this as $(-2{)}^8\cdot i^8$. Since $8$ is an even power, the negative sign disappears, and $2^8=256$.

For $i^8$, we know $i^4=1$, so $i^8=(i^4{)}^2=1^2=1$. The final result is: