Problem 27 - Precalculus with Limits

Question

Represent the complex number graphically, and find the trigonometric form of the number. $$2 \sqrt{2}-i$$

Step-by-Step Solution

Answer

The complex number \(2\sqrt{2} - i\) can be represented graphically in the 4th quadrant with a magnitude of 3 and a trigonometric form, \( 3( \cos (2\pi - \arctan(\frac{1}{2\sqrt{2}}))+i \sin(2\pi - \arctan(\frac{1}{2\sqrt{2}}))) \)

Step 1: Graphical Representation

Firstly, plot the real part on the x-axis and the imaginary part on the y-axis. So, \(2 \sqrt{2}\) will be plotted on the x-axis and \(-1\) on the y-axis.

Step 2: Calculate Magnitude (r)

The magnitude, r, of the complex number is the distance from the origin to the point. This can be calculated using the Pythagorean theorem by squaring the real and imaginary parts, adding them together, and then finding the square root of the sum. \(r =\sqrt{(2 \sqrt{2})^2 + (-1)^2} = \sqrt{8 + 1} = 3\)

Step 3: Calculate the Angle (\( \theta \))

Next, calculate the argument or the angle \( \theta \) which the line makes with the positive x-axis in counter-clockwise direction. This can be done using the inverse tangent function. However, pay attention to each quadrant as each has a different formula for \( \theta \). Our number in the fourth quadrant (negative imaginary part), so we use the formula \( \theta = 2\pi - \arctan(y/x)\). Putting values in the formula, we get, \( \theta = 2\pi - \arctan(\frac{1}{2 \sqrt{2}})\)

Step 4: Trigonometric Form

After finding r (the magnitude) and \( \theta \) (the angle), we can now represent the complex number in trigonometric form. The trigonometric form of the complex number is \(r(\cos{\theta}+i\sin{\theta})\). Substituting our values, we get \(3(\cos(\theta)+i\sin(\theta))=\allowbreak 3[ \cos ( 2\pi - \arctan(\frac{1}{2\sqrt{2}}))+ i \sin(2\pi - \arctan(\frac{1}{2\sqrt{2}}))] \)

Key Concepts

Graphical Representation of Complex Numbers

Understanding the graphical representation of complex numbers helps to visualize them easily. A complex number has a real part and an imaginary part. To graphically represent a complex number like \(2\sqrt{2} - i\), plot its components on a coordinate plane. The real part \(2\sqrt{2}\) goes along the horizontal x-axis and the imaginary part \(-1\) is placed on the vertical y-axis. Therefore, the point you plot is \((2\sqrt{2}, -1)\). This graphical approach makes it simpler to understand complex numbers, their relationships, and calculations that involve them.

Magnitude of Complex Numbers

The magnitude of a complex number gives us an idea of how far the number is from the origin on the coordinate plane. It is also known as the modulus or the absolute value. For a complex number \(a + bi\), the magnitude \(r\) can be calculated using the Pythagorean theorem: \(r = \sqrt{a^2 + b^2}\).

In the case of \(2\sqrt{2} - i\), the magnitude calculation involves: \(r = \sqrt{(2\sqrt{2})^2 + (-1)^2} = \sqrt{8 + 1} = 3\). Remember, the magnitude is always a non-negative real number and it’s the distance from the origin to the point on the graph.

Argument of Complex Numbers

The argument of a complex number is the angle formed between the positive real axis and the line connecting the origin to the point representing the complex number. For \(2\sqrt{2} - i\), since it lies in the fourth quadrant (as the imaginary part is negative), the argument is calculated using: \(\theta = 2\pi - \arctan\left(\frac{1}{2\sqrt{2}}\right)\).

It's crucial to review which quadrant the complex number lies in because this affects the calculation of its argument. The argument is measured in radians and helps in converting the complex number into its polar or trigonometric form.

Pythagorean Theorem in Complex Numbers

The Pythagorean theorem is a valuable tool when working with complex numbers because it allows us to determine the magnitude. It states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

• For complex numbers, this is applied to calculate the magnitude, where the real part is one side and the imaginary part is the other.
• For example, with \(2\sqrt{2} - i\), the measures \((2\sqrt{2})^2 = 8\) for the real part and \((-1)^2 = 1\) for the imaginary part add up.

This gives us \(\sqrt{8 + 1} = 3\), representing the magnitude. Using this theorem simplifies complex calculations and enhances understanding of relationships in the complex plane.