Problem 35 - Precalculus

Question

Use a calculator to express each complex number in polar form. $$5.1+2.3 i$$

Step-by-Step Solution

Answer

The polar form is approximately \(5.59(\cos 24.23^{\circ} + i \sin 24.23^{\circ})\).

Step 1: Find the Magnitude

The magnitude (or modulus) of a complex number \(a + bi\) is given by \(\sqrt{a^2 + b^2}\). For the complex number \(5.1 + 2.3i\):\[|z| = \sqrt{(5.1)^2 + (2.3)^2}\]Calculating this, we have:\[|z| = \sqrt{26.01 + 5.29} = \sqrt{31.3} \approx 5.59\]

Step 2: Calculate the Argument

The argument \(\theta\) of a complex number is given by \(\tan^{-1}(\frac{b}{a})\). For \(5.1 + 2.3i\), the argument is:\[\theta = \tan^{-1}\left(\frac{2.3}{5.1}\right)\]Using a calculator, this gives:\[\theta \approx \tan^{-1}(0.451) \approx 24.23^{\circ}\]

Step 3: Express in Polar Form

The polar form of a complex number is \(r(\cos \theta + i \sin \theta)\) or \(r\text{cis}\,\theta\). Here, \(r = 5.59\) and \(\theta = 24.23^{\circ}\):\[5.1 + 2.3i = 5.59(\cos 24.23^{\circ} + i \sin 24.23^{\circ})\]

Key Concepts

Polar Form

Complex numbers can be represented in different forms, and one such form is the polar form. Polar form provides a way to express complex numbers using a magnitude and angle, rather than a real and imaginary part. To convert a complex number like \(5.1 + 2.3i\) into polar form, we first determine its magnitude and argument. In polar form, a complex number \(a + bi\) is expressed as \(r(\cos \theta + i \sin \theta)\) or \(r\text{cis}\,\theta\), where \(r\) is the magnitude and \(\theta\) is the argument. This representation is especially useful in fields like engineering, physics, and computer science as it simplifies multiplication and division of complex numbers.

Magnitude of Complex Numbers

The magnitude of a complex number \(a + bi\) is calculated using the formula \(\sqrt{a^2 + b^2}\). It's essentially the distance of the complex number from the origin in the complex plane. For the complex number \(5.1 + 2.3i\), its magnitude is given by:

• Compute \(a^2\): \(5.1^2 = 26.01\)
• Compute \(b^2\): \(2.3^2 = 5.29\)
• Add \(a^2 + b^2\): \(26.01 + 5.29 = 31.3\)
• Take the square root: \(\sqrt{31.3} \approx 5.59\)

The result, \(5.59\), is the magnitude, representing how far \(5.1 + 2.3i\) is from the origin.

Argument of Complex Numbers

The argument of a complex number is the angle formed with the positive real axis in the complex plane. It can be computed from a complex number \(a + bi\) as \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\). For \(5.1 + 2.3i\):

• Divide \(b\) by \(a\): \(\frac{2.3}{5.1} \approx 0.451\)
• Calculate the inverse tangent: \(\theta = \tan^{-1}(0.451)\)
• Using a calculator, \(\theta \approx 24.23^{\circ}\)

This angle, approximately \(24.23^{\circ}\), is the argument, indicating the direction of the complex number in the plane.

Trigonometry

Trigonometry is integral to understanding complex numbers in polar form. We use trigonometric functions like cosine and sine to represent the components of complex numbers in this format. In

• Polar form uses \(\cos \theta\) and \(\sin \theta\) to express the horizontal and vertical components, respectively, of the vector representing a complex number.
• By configuring complex numbers with trigonometric expressions, we simplify many operations, including transformation and rotation of vectors.
• In this exercise, the trigonometry is seen when representing the complex number as \(5.59(\cos 24.23^{\circ} + i \sin 24.23^{\circ})\).

Such representations allow for a more geometric understanding of complex numbers, closely tying to concepts of angles and radii in circles.

Precalculus

Precalculus provides the foundational understanding of complex numbers, readying students for more advanced mathematical studies.

• This subject connects algebraic and geometric interpretations, empowering students to convert complex numbers between rectangular (\(a + bi\)) and polar forms.
• Key functions like magnitudes, arguments, and trigonometric operations form precalculus' toolbox, enabling them to handle complex number tasks like the one shown in this exercise.
• Through this conversion, precalculus sets the stage for clear visual and computational solutions, making it crucial for tackling more sophisticated concepts in calculus and beyond.

Understanding these concepts is not just about solving equations, but also about visualizing mathematical ideas effectively.