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Chapter 8: Problem 16

In Exercises 13-28, express each complex number in polar form. $$ -3-\sqrt{3} i $$

Short Answer

Expert verified
Polar form is \(2\sqrt{3} e^{i\frac{7\pi}{6}}\).

Step by step solution

01

Understand the Cartesian Form

The given complex number is \(-3 - \sqrt{3} i\). This is in Cartesian form \(a + bi\), where \(a = -3\) and \(b = -\sqrt{3}\).
02

Calculate the Magnitude

The magnitude \(r\) of a complex number \(a + bi\) is calculated using the formula: \[r = \sqrt{a^2 + b^2}.\] Substituting the values \(a = -3\) and \(b = -\sqrt{3}\): \[r = \sqrt{(-3)^2 + (-\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}.\]
03

Determine the Angle

The angle \( \theta \) in polar coordinates is found using the tangent function: \[\theta = \arctan\left(\frac{b}{a}\right).\] For \(a = -3\) and \(b = -\sqrt{3}\), the angle is: \[\theta = \arctan\left(\frac{-\sqrt{3}}{-3}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right).\] Since the complex number lies in the third quadrant (both components are negative), we need to add \(\pi\) to the angle found in the first quadrant: \[\theta = \frac{\pi}{6} + \pi = \frac{7\pi}{6}.\]
04

Write in Polar Form

The polar form of a complex number is expressed as \[r(\cos \theta + i\sin \theta),\] which can also be written using Euler's formula as \[r e^{i\theta}.\] Substituting \(r = 2\sqrt{3}\) and \(\theta = \frac{7\pi}{6}\): \[-3 - \sqrt{3} i = 2\sqrt{3} e^{i\frac{7\pi}{6}}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Number
Complex numbers are the foundation of many areas in mathematics and engineering. They extend the idea of real numbers by incorporating an imaginary unit, denoted as \(i\), where \(i^2 = -1\). This allows complex numbers to be expressed in the form \(a + bi\). Here, \(a\) is the real part and \(b\) is the imaginary part of the complex number.

Complex numbers are incredibly useful since they allow for the computation and representation of square roots of negative numbers, which cannot be handled by real numbers alone. Complex numbers can be added, subtracted, multiplied, and divided much like real numbers, with the additional consideration of handling the imaginary unit \(i\).
  • Real component: \(a\)
  • Imaginary component: \(bi\)
Cartesian Form
The Cartesian form of a complex number refers to its standard expression as \(a + bi\). This notation is derived from Cartesian coordinates in a plane, where "\(a\)" serves as the horizontal axis value, and "\(b\)" represents the vertical axis value.

In this form, the complex number can be easily visualized as a point on the complex plane, with the horizontal axis serving as the real part and the vertical axis as the imaginary part. This spatial representation makes it simple to understand both the magnitude and direction of complex numbers.
  • Easy visualization on the complex plane.
  • Simplifies addition and subtraction of complex numbers.
The Cartesian form is important for calculations and is often used as the starting point for converting to other forms, like polar form.
Magnitude of Complex Numbers
The magnitude of a complex number, sometimes referred to as the modulus, is a measure of its size or distance from the origin in the complex plane. Given a complex number \(a + bi\), the magnitude \(r\) is found using the formula:\[r = \sqrt{a^2 + b^2}.\]

In our exercise, the magnitude of the complex number \(-3 - \sqrt{3}i\) is calculated as \(2\sqrt{3}\). This tells us how far the point represented by the complex number is from the origin, independent of its direction. Calculating the magnitude is a key step in converting a complex number from Cartesian form to polar form, as it provides the radial coordinate in polar coordinates.
Angle in Polar Coordinates
To transition from Cartesian to polar form, determining the angle \(\theta\) is necessary. The angle gives the direction of the vector representing the complex number relative to the positive real axis.

For a complex number \(a + bi\), the angle \(\theta\) is calculated using the arctangent function:\[\theta = \arctan\left(\frac{b}{a}\right).\]
Understanding which quadrant the complex number lies in is crucial since it affects the angle's calculation. For example, if both \(a\) and \(b\) are negative, the complex number is in the third quadrant, requiring an adjustment of the calculated angle by adding \(\pi\). In our exercise, the angle \(\theta = \frac{7\pi}{6}\) is derived by considering the initial angle \(\frac{\pi}{6}\) and adjusting for its position in the third quadrant.
  • Helps express complex numbers in polar form.
  • Provides the angular coordinate in polar coordinates.

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Most popular questions from this chapter

In Exercises 45-68, graph each equation. In Exercises 63-68, convert the equation from polar to rectangular form first and identify the resulting equation as a line, parabola, or circle. $$ r^{2} \cos ^{2} \theta-2 r \cos \theta+r^{2} \sin ^{2} \theta=8 $$

Consider the parametric equations \(x=\cos (a t)\) and \(y=\sin (b t)\). Use a graphing utility to explore the graphs for \(a=2\) and \(b=4, a=4\) and \(b=2, a=1\) and \(b=3\), and \(a=3\) and \(b=1\). Find the \(t\)-interval that gives one cycle of the curve.

For Exercises 37-46, recall that the flight of a projectile can be modeled with the parametric equations $$ x=\left(v_{0} \cos \theta\right) t \quad y=-16 t^{2}+\left(v_{0} \sin \theta\right) t+h $$ where \(t\) is in seconds, \(v_{0}\) is the initial velocity in feet per second, \(\theta\) is the initial angle with the horizontal, and \(h\) is the initial height above ground, where \(x\) and \(y\) are in feet. Flight of a Projectile. A projectile is launched from the ground at a speed of 400 feet per second at an angle of \(45^{\circ}\) with the horizontal. After how many seconds does the projectile hit the ground?

In Exercises 1-20, graph the curve defined by the following sets of parametric equations. Be sure to indicate the direction of movement along the curve. $$ x=4 \cos (2 t), y=t, t \text { in }[0,2 \pi] $$

Bicycle Racing. A boy on a bicycle racing around an oval track has a position given by the equations \(x=-100 \sin \left(\frac{t}{4}\right)\) and \(y=75 \cos \left(\frac{t}{4}\right)\), where \(x\) and \(y\) are the horizontal and vertical positions in feet relative to the center of the track \(t\) seconds after the start of the race. Find the boy's position at \(t=10,20\), and 30 .
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