Finding the Modulus of a Complex Number Raised to a Power

The question asks for the modulus of the complex number \(\left(\frac{\sqrt{-3}}{2}-\frac{1}{2}\right)^{200}\).

First, let's simplify the base of the expression, which is \(\frac{\sqrt{-3}}{2}-\frac{1}{2}\).

We know that \(\sqrt{-3} = \sqrt{3} \times \sqrt{-1} = i\sqrt{3}\), where \(i\) is the imaginary unit (\(i^2 = -1\)).

So, the complex number base becomes:

\(z = \frac{i\sqrt{3}}{2} - \frac{1}{2} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\)

To find the modulus of a complex number raised to a power, a key property of complex numbers is very useful: \(|z^n| = |z|^n\). This means the modulus of a complex number raised to a power is equal to the modulus of the complex number raised to that same power.

Let's find the modulus of the base complex number, \(z = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\).

For a complex number \(z = x + iy\), the modulus \(|z|\) is calculated as \(|z| = \sqrt{x^2 + y^2}\).

In this case, \(x = -\frac{1}{2}\) and \(y = \frac{\sqrt{3}}{2}\).

The modulus of \(z\) is:

\(|z| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\)

\(|z| = \sqrt{\frac{1}{4} + \frac{3}{4}}\)

\(|z| = \sqrt{\frac{1+3}{4}}\)

\(|z| = \sqrt{\frac{4}{4}}\)

\(|z| = \sqrt{1}\)

\(|z| = 1\)

Now, we need to find the modulus of \(z^{200}\), where \(z = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\).

Using the property \(|z^n| = |z|^n\), we have:

\(\left|\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^{200}\right| = \left|-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right|^{200}\)

We calculated the modulus of the base to be 1.

So, the modulus of the expression is \(1^{200}\).

\(1^{200} = 1\)

Thus, the modulus of \(\left(\frac{\sqrt{-3}}{2}-\frac{1}{2}\right)^{200}\) is 1.

This base complex number \(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\) is actually a special complex number, often denoted as \(\omega\), which is one of the complex cube roots of unity (excluding 1).

In polar form, \(-\frac{1}{2} + i\frac{\sqrt{3}}{2} = 1 \cdot \left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) = e^{i\frac{2\pi}{3}}\). Its modulus is 1.

Using De Moivre's theorem, \(z^n = r^n(\cos(n\theta) + i\sin(n\theta))\), where \(z = r(\cos\theta + i\sin\theta)\). The modulus of \(z^n\) is \(r^n\).

Here, \(r=1\) and \(n=200\). So the modulus is \(1^{200} = 1\).

Both methods confirm the result.

The final answer is 1.

Revision Table: Key Complex Number Concepts

Additional Information on Complex Number Properties

Understanding the properties of complex numbers is crucial for solving problems involving powers and roots. The property \(|z^n| = |z|^n\) simplifies calculations significantly, especially when dealing with large exponents like 200 in this question. Instead of calculating the complex number \(\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^{200}\) first and then finding its modulus, we can directly find the modulus of the base and raise it to the power 200.

The complex number \(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\) is a special case. Its argument is \(120^\circ\) or \(\frac{2\pi}{3}\) radians, and its modulus is 1. Complex numbers with a modulus of 1 lie on the unit circle in the complex plane. When such a complex number is raised to a power, its modulus remains 1, while its argument is multiplied by the power, as described by De Moivre's Theorem.

In this problem, \(z = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\). In polar form, \(z = 1 \cdot e^{i\frac{2\pi}{3}}\). Then \(z^{200} = (e^{i\frac{2\pi}{3}})^{200} = e^{i\frac{400\pi}{3}}\). The modulus of \(e^{i\theta}\) is always 1, regardless of the value of \(\theta\). So, \(|e^{i\frac{400\pi}{3}}| = 1\). This confirms the result using the polar form and Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\).