Problem 23
Question
Express each complex number in polar form. $$-5+5 i$$
Step-by-Step Solution
Verified Answer
Polar form: \( 5\sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \).
1Step 1: Identify the Real and Imaginary Parts
In this complex number, the real part is \(-5\) and the imaginary part is \(5\). These are the coordinates that will be used in the polar form calculation.
2Step 2: Calculate the Magnitude (r)
The magnitude \( r \) of a complex number \( a + bi \) is calculated using the formula \( r = \sqrt{a^2 + b^2} \). For \(-5+5i\), calculate as follows: \[ r = \sqrt{(-5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]
3Step 3: Determine the Argument (θ)
The argument \( \theta \) is the angle formed with the positive real axis. It's calculated using \( \tan \theta = \frac{b}{a} \). Here, \( \tan \theta = \frac{5}{-5} = -1 \). Since the point \((-5, 5)\) is in the second quadrant, and \( \tan \theta = -1 \) is consistent with an angle of \( 3\pi/4 \) radians, \( \theta = 3\pi/4 \).
4Step 4: Write the Polar Form
The polar form of a complex number is expressed as \( r(\cos \theta + i \sin \theta) \). Using \( r = 5\sqrt{2} \) and \( \theta = 3\pi/4 \) from previous steps, the polar form is: \[ 5\sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \]
Key Concepts
polar formmagnitudeargumentsecond quadrant
polar form
The polar form of a complex number is a fascinating way to represent complex numbers, leveraging both magnitude and direction. It's quite like polar coordinates in geometry. In polar form, a complex number isn't expressed using the usual way, like \'a + bi\'. Instead, it's represented as \('r \cdot \cos(\theta) + i\cdot \sin(\theta)'\). Here, \'r\' stands for the magnitude (or the length) of the vector representing the complex number. On the other hand, \'\theta\' is the argument, which is the angle that the line makes with the positive real axis. This way of representation gives an insight into both the size and direction of the complex number.
There are a couple of advantages when using polar form:
There are a couple of advantages when using polar form:
- It makes multiplication and division of complex numbers easier to visualize and perform.
- It clearly separates the magnitude and direction aspects of a complex number.
magnitude
Understanding the magnitude of a complex number is the key to appreciating its size, much like finding the length of a vector in geometry. The magnitude, often denoted by \'r\' or \'|z|\', represents the distance of the complex number from the origin when plotted on the complex plane. A very useful formula to find the magnitude of a complex number \'a + bi\' is \('r = \sqrt{a^2 + b^2}'\).
Let's break it down using our example:- Real part: \(-5\)- Imaginary part: \(+5\)By substituting these into the formula, we calculate: \( r = \sqrt{(-5)^2 + (5)^2} = \sqrt{50} = 5\sqrt{2} \)This implies that the magnitude is \('5\sqrt{2}'\), conveying that the length of the vector representing \(-5 + 5i\) from the origin is \('5\sqrt{2}'\) units.
Let's break it down using our example:- Real part: \(-5\)- Imaginary part: \(+5\)By substituting these into the formula, we calculate: \( r = \sqrt{(-5)^2 + (5)^2} = \sqrt{50} = 5\sqrt{2} \)This implies that the magnitude is \('5\sqrt{2}'\), conveying that the length of the vector representing \(-5 + 5i\) from the origin is \('5\sqrt{2}'\) units.
argument
The argument of a complex number is essential when understanding its direction on the complex plane; it tells us the angle the vector makes with the positive real axis. In mathematical terms, it's calculated using the formula \( \tan \theta = \frac{b}{a} \).
For our complex number \( -5 + 5i \), this results in \( \tan \theta = \frac{5}{-5} = -1 \). Since we're dealing with a point in the second quadrant, we align with typical angles associated with tangent values. Known angles that give \( \tan \theta = -1 \) can be \( \theta = 3\pi/4 \ \text{or} \ \theta = 7\pi/4\), depending on the quadrant. Therefore, for the point \( (-5, 5) \), the angle is \( \theta = 3\pi/4 \). Understanding the argument helps in visualizing the kind of rotation from the positive real axis needed to reach this complex number's direction.
For our complex number \( -5 + 5i \), this results in \( \tan \theta = \frac{5}{-5} = -1 \). Since we're dealing with a point in the second quadrant, we align with typical angles associated with tangent values. Known angles that give \( \tan \theta = -1 \) can be \( \theta = 3\pi/4 \ \text{or} \ \theta = 7\pi/4\), depending on the quadrant. Therefore, for the point \( (-5, 5) \), the angle is \( \theta = 3\pi/4 \). Understanding the argument helps in visualizing the kind of rotation from the positive real axis needed to reach this complex number's direction.
second quadrant
The concept of quadrants is vital for determining the position of points on the complex plane. As we remember from geometry, the complex plane can be divided into four quadrants using the real and imaginary axes. These sections help in identifying the sign of the real and imaginary components.
The second quadrant falls between \( \pi/2 \) and \( \pi \) on the unit circle, and it is characterized by:
This information is crucial when calculating the argument since the angle must reflect this quadrant's properties.
The second quadrant falls between \( \pi/2 \) and \( \pi \) on the unit circle, and it is characterized by:
- Negative real parts
- Positive imaginary parts
This information is crucial when calculating the argument since the angle must reflect this quadrant's properties.
Other exercises in this chapter
Problem 23
Convert each point to exact rectangular coordinates. $$\left(-1, \frac{5 \pi}{6}\right)$$
View solution Problem 23
Find the result of each expression using De Moivre's theorem. Write the answer in rectangular form. $$(-\sqrt{3}+i)^{6}$$
View solution Problem 23
Perform the indicated vector operation, given \(u=(-4,3)\) and \(v=\langle 2,-5\rangle\) $$6(\mathbf{u}-\mathbf{v})$$
View solution Problem 24
Find the angle (round to the nearest degree) between each pair of vectors. $$\langle 2,8\rangle \text { and }\langle-12,3\rangle$$
View solution