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Magazine Chapter 7: Problem 20
Express each complex number in polar form. $$\frac{1}{6}-\frac{1}{6} i$$
Short Answer
Expert verified
Polar form is \( \frac{1}{3 \sqrt{2}} e^{-i\frac{\pi}{4}} \).
Step by step solution
01
Identify the Real and Imaginary Parts
First, identify the real and imaginary parts of the complex number. The complex number given is \( \frac{1}{6} - \frac{1}{6} i \). Here, the real part \( a \) is \( \frac{1}{6} \) and the imaginary part \( b \) is \( -\frac{1}{6} \).
02
Calculate the Magnitude
The magnitude (or modulus) \( r \) of a complex number \( a + bi \) is given by \( r = \sqrt{a^2 + b^2} \). Substitute \( a = \frac{1}{6} \) and \( b = -\frac{1}{6} \) into the formula:\[r = \sqrt{\left(\frac{1}{6}\right)^2 + \left(-\frac{1}{6}\right)^2} = \sqrt{\frac{1}{36} + \frac{1}{36}} = \sqrt{\frac{2}{36}} = \sqrt{\frac{1}{18}} = \frac{1}{\sqrt{18}} = \frac{1}{3 \sqrt{2}}\]
03
Calculate the Argument
The argument \( \theta \) is found using \( \theta = \text{atan2}(b, a) \). With \( a = \frac{1}{6} \) and \( b = -\frac{1}{6} \), we calculate:\[\theta = \text{atan2}\left(-\frac{1}{6}, \frac{1}{6}\right) = \text{atan2}(-1, 1) = -\frac{\pi}{4} \text{ radians}\]Here, \( \text{atan2} \) automatically takes into account the sign of both arguments and gives the angle in the correct quadrant.
04
Express in Polar Form
Using the magnitude \( r \) and the argument \( \theta \), express the complex number in polar form: \( r(\cos \theta + i\sin \theta) \) or simply \( re^{i\theta} \).Substitute the calculated values:\[\text{Polar Form} = \frac{1}{3 \sqrt{2}} \left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)\]Or in exponential form:\[= \frac{1}{3 \sqrt{2}} e^{-i\frac{\pi}{4}}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnitude of Complex Number
The magnitude of a complex number is essential when dealing with its polar form. Imagine a complex number as a point in a two-dimensional space, with the real part on the x-axis and the imaginary part on the y-axis. The magnitude, also known as the modulus, is the distance of that point from the origin of this plane. To find the magnitude, use the Pythagorean theorem in the form:\[ r = \sqrt{a^2 + b^2} \]where \( a \) is the real part and \( b \) is the imaginary part of the complex number.
- For the number \( \frac{1}{6} - \frac{1}{6}i \), both real and imaginary parts are \( \frac{1}{6} \) and \( -\frac{1}{6} \) respectively.
- Substituting these values into our formula provides: \[ r = \sqrt{\left(\frac{1}{6}\right)^2 + \left(-\frac{1}{6}\right)^2} = \sqrt{\frac{2}{36}} = \frac{1}{3\sqrt{2}} \]
Argument of Complex Number
The argument of a complex number gives the angle of the line from the origin to the point representing the number in the complex plane. This angle is measured from the positive x-axis (real axis) towards the line, rotating counter-clockwise. To find the argument, we use the function \( \text{atan2}(b, a) \), where \( a \) and \( b \) are the real and imaginary components, respectively.For our example, with \( a = \frac{1}{6} \) and \( b = -\frac{1}{6} \), the argument is calculated as:\[\theta = \text{atan2}\left(-\frac{1}{6}, \frac{1}{6}\right) = \text{atan2}(-1, 1) = -\frac{\pi}{4}\]Some points to keep in mind:
- The \( \text{atan2} \) function considers the sign of both \( a \) and \( b \), placing the argument \( \theta \) in the correct quadrant of the plane.
- An angle of \( -\frac{\pi}{4} \) radians indicates that the direction is \( 45 \) degrees below the positive x-axis.
Complex Number Conversion
Turning a complex number into its polar form is like switching from one language to another, but still describing the same object. Instead of using a Cartesian form \( a + bi \), we use polar coordinates, which consist of a magnitude \( r \) and an angle \( \theta \). The conversion is expressed in polar notation as:\[ r(\cos \theta + i\sin \theta) \quad\text{or}\quad re^{i\theta}\]Here is how you can express the complex number \( \frac{1}{6} - \frac{1}{6}i \) in polar form:
- First, we found the magnitude \( r = \frac{1}{3 \sqrt{2}} \).
- Next, we calculated the argument \( \theta = -\frac{\pi}{4} \).
- Combining these in polar form gives: \[ \frac{1}{3 \sqrt{2}}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right) \quad \text{or} \quad \frac{1}{3 \sqrt{2}} e^{-i\frac{\pi}{4}} \]
