Finding Modulus And Argument Of Complex Number Z = [√3(cos(π/6) + I Sin(π/6))]^4 / (cos(π/4) - I Sin(π/4))^2

Hey there, math enthusiasts! Today, we're going to dive into the fascinating world of complex numbers and tackle a problem that might seem daunting at first glance. But fear not, we'll break it down step by step and make sure you grasp the concepts involved. Our mission? To find the modulus and argument of the complex number:

$$z=\frac{\left[\sqrt{3}\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\right]^4}{\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)^2}$$

So, buckle up and let's embark on this mathematical journey together!

Decoding the Complex Number: Modulus and Argument Explained

Before we jump into the calculations, let's take a moment to understand what the modulus and argument of a complex number actually represent. In essence, they provide us with a way to visualize complex numbers in a geometric context. Think of the complex plane, where the horizontal axis represents the real part and the vertical axis represents the imaginary part of a complex number. Any complex number can be plotted as a point on this plane.

The modulus, often denoted as |z|, is the distance from the origin (0, 0) to the point representing the complex number in the complex plane. It's essentially the magnitude or absolute value of the complex number. The modulus is always a non-negative real number.

The argument, denoted as arg(z), is the angle formed between the positive real axis and the line segment connecting the origin to the point representing the complex number in the complex plane. This angle is measured in radians and is considered positive for counterclockwise rotations and negative for clockwise rotations. The argument is not unique, as adding multiples of 2π to the angle will result in the same complex number. Therefore, we often consider the principal argument, which lies in the interval (-π, π].

Understanding the modulus and argument is crucial because they allow us to express complex numbers in polar form, which simplifies many calculations, especially when dealing with powers and roots of complex numbers. The polar form of a complex number z is given by:

$$z = r(\cos θ + i \sin θ)$$

where r is the modulus |z| and θ is the argument arg(z). This representation beautifully connects the algebraic and geometric aspects of complex numbers.

Step-by-Step Solution: Unraveling the Complex Expression

Now that we have a solid grasp of the concepts, let's tackle the given complex number step by step. Our goal is to simplify the expression and then determine its modulus and argument.

1. Simplifying the Numerator: Powering Up the Complex Term

Let's first focus on the numerator of our complex number:

$$\left[\sqrt{3}\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\right]^4$$

We can use De Moivre's Theorem to simplify this expression. De Moivre's Theorem states that for any complex number in polar form, z = r(cos θ + i sin θ), and any integer n:

$$z^n = r^n(\cos(nθ) + i \sin(nθ))$$

Applying this theorem to our numerator, we get:

$$\left[\sqrt{3}\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\right]^4 = (\sqrt{3})^4 \left(\cos \left(4 \cdot \frac{\pi}{6}\right)+i \sin \left(4 \cdot \frac{\pi}{6}\right)\right)$$

Simplifying the terms:

$$= 9 \left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)$$

2. Simplifying the Denominator: Taming the Complex Expression

Next, let's simplify the denominator:

$$\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)^2$$

Here, we encounter a slight twist: the imaginary part has a negative sign. To apply De Moivre's Theorem directly, we need to express the complex number in the standard polar form with a positive imaginary term. We can achieve this by recognizing that:

$$\cos(-\theta) = \cos(\theta) \quad \text{and} \quad \sin(-\theta) = -\sin(\theta)$$

Therefore, we can rewrite the denominator as:

$$\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)^2 = \left(\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right)^2$$

Now, we can apply De Moivre's Theorem:

$$= \left(\cos \left(2 \cdot \left(-\frac{\pi}{4}\right)\right)+i \sin \left(2 \cdot \left(-\frac{\pi}{4}\right)\right)\right)$$

Simplifying:

$$= \cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right) = 0 - i = -i$$

3. Combining Numerator and Denominator: The Grand Finale

Now that we've simplified both the numerator and the denominator, we can put them together:

$$z = \frac{9 \left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)}{-i}$$

To further simplify this expression, we need to get rid of the complex number in the denominator. We can do this by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of -i is simply i:

$$z = \frac{9 \left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)}{-i} \cdot \frac{i}{i} = \frac{9i \left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)}{-i^2}$$

Since i² = -1, we have:

$$z = 9i \left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)$$

Now, let's distribute the 9i:

$$z = 9i \cos \frac{2\pi}{3} + 9i^2 \sin \frac{2\pi}{3} = 9i \cos \frac{2\pi}{3} - 9 \sin \frac{2\pi}{3}$$

We know that cos(2π/3) = -1/2 and sin(2π/3) = √3/2. Substituting these values:

$$z = 9i \left(-\frac{1}{2}\right) - 9 \left(\frac{\sqrt{3}}{2}\right) = -\frac{9\sqrt{3}}{2} - \frac{9}{2}i$$

4. Finding the Modulus: Measuring the Magnitude

Now that we have the complex number in the standard form a + bi, we can find its modulus using the formula:

$$|z| = \sqrt{a^2 + b^2}$$

In our case, a = -9√3/2 and b = -9/2. Plugging these values into the formula:

$$|z| = \sqrt{\left(-\frac{9\sqrt{3}}{2}\right)^2 + \left(-\frac{9}{2}\right)^2} = \sqrt{\frac{243}{4} + \frac{81}{4}} = \sqrt{\frac{324}{4}} = \sqrt{81} = 9$$

So, the modulus of our complex number is 9.

5. Finding the Argument: Unveiling the Angle

To find the argument, we use the formula:

$$arg(z) = \arctan\left(\frac{b}{a}\right)$$

However, we need to be careful about the quadrant in which the complex number lies. Our complex number z = -9√3/2 - (9/2)i has both a negative real part and a negative imaginary part, which means it lies in the third quadrant. The arctangent function will give us an angle in the first or fourth quadrant, so we need to adjust it accordingly.

Let's first calculate the arctangent:

$$\arctan\left(\frac{b}{a}\right) = \arctan\left(\frac{-\frac{9}{2}}{-\frac{9\sqrt{3}}{2}}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$$

Since our complex number is in the third quadrant, we need to add -π to this angle to get the principal argument:

$$arg(z) = \frac{\pi}{6} - \pi = -\frac{5\pi}{6}$$

Therefore, the argument of our complex number is -5π/6.

Conclusion: Mastering Complex Numbers

And there you have it! We've successfully navigated the complex expression, simplified it, and found both its modulus and its argument. To recap, the modulus of the complex number is 9, and its argument is -5π/6.

This exercise highlights the power of De Moivre's Theorem and the importance of understanding the geometric interpretation of complex numbers. By breaking down the problem into smaller, manageable steps, we were able to conquer what initially seemed like a complex challenge.

So, keep practicing, keep exploring, and you'll become a master of complex numbers in no time! Remember, math is not just about formulas and equations; it's about understanding the underlying concepts and developing a problem-solving mindset. Keep up the great work, guys!