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Magazine Chapter 8: Problem 59
Write \(z_{1}\) and \(z_{2}\) in polar form, and then find the product \(z_{1} z_{2}\) and the quotients \(z_{1} / z_{2}\) and \(1 / z_{1}\). $$z_{1}=2 \sqrt{3}-2 i, \quad z_{2}=-1+i$$
Short Answer
Expert verified
In polar form, the product is \(4\sqrt{2}\left(\cos\frac{7\pi}{12} + i\sin\frac{7\pi}{12}\right)\), \(\frac{z_1}{z_2} = 2\sqrt{2}(\cos(-\frac{13\pi}{12}) + i\sin(-\frac{13\pi}{12}))\), and \(\frac{1}{z_1} = \frac{1}{4} (\cos\frac{\pi}{6} + i\sin\frac{\pi}{6})\).
Step by step solution
01
Convert to Polar Form: For \( z_1 \)
To convert the complex number \( z_1 = 2\sqrt{3} - 2i \) into polar form, first calculate the modulus \( |z_1| = \sqrt{(2\sqrt{3})^2 + (-2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4 \). Then, find the argument \( \theta = \tan^{-1}\left(\frac{-2}{2\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \). Thus, \( z_1 = 4 \left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right) \).
02
Convert to Polar Form: For \( z_2 \)
For the complex number \( z_2 = -1 + i \), calculate the modulus \( |z_2| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \). Then find the argument \( \theta = \tan^{-1}\left(\frac{1}{-1}\right) = \tan^{-1}(-1) = \frac{3\pi}{4} \). Thus, \( z_2 = \sqrt{2} \left(\cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)\right) \).
03
Find the Product \( z_1 z_2 \)
The product \( z_1 z_2 \) in polar form is found by multiplying the moduli and adding the arguments: \( |z_1 z_2| = 4 \times \sqrt{2} = 4\sqrt{2} \), and the argument is \( -\frac{\pi}{6} + \frac{3\pi}{4} = \frac{7\pi}{12} \). So \( z_1 z_2 = 4\sqrt{2} \left( \cos\left( \frac{7\pi}{12} \right) + i\sin\left( \frac{7\pi}{12} \right) \right) \).
04
Find the Quotient \( \frac{z_1}{z_2} \)
For the quotient \( \frac{z_1}{z_2} \), divide the moduli and subtract the arguments: \( \left|\frac{z_1}{z_2}\right| = \frac{4}{\sqrt{2}} = 2\sqrt{2} \), and the argument is \( -\frac{\pi}{6} - \frac{3\pi}{4} = -\frac{13\pi}{12} \). Thus, \( \frac{z_1}{z_2} = 2\sqrt{2} \left( \cos\left( -\frac{13\pi}{12} \right) + i\sin\left( -\frac{13\pi}{12} \right) \right) \).
05
Find the Quotient \( \frac{1}{z_1} \)
The reciprocal \( \frac{1}{z_1} \) involves finding the reciprocal of the modulus and negating the argument: \( \left|\frac{1}{z_1}\right| = \frac{1}{4} \), and the argument is \( \frac{\pi}{6} \). Therefore, \( \frac{1}{z_1} = \frac{1}{4} \left( \cos\left( \frac{\pi}{6} \right) + i\sin\left( \frac{\pi}{6} \right) \right) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Form
Converting complex numbers to polar form is a powerful way to understand and perform operations with them. In polar form, a complex number is expressed using its modulus and argument. This means that any complex number \( z \) can be represented as \( z = r(\cos \theta + i \sin \theta) \), where \( r \) is the modulus and \( \theta \) is the argument.
To convert a complex number \( z = a + bi \) to polar form, follow these steps:
To convert a complex number \( z = a + bi \) to polar form, follow these steps:
- Calculate the modulus: \( r = \sqrt{a^2 + b^2} \).
- Determine the argument: \( \theta = \tan^{-1}(\frac{b}{a}) \).
Modulus
The modulus of a complex number provides the measure of its distance from the origin in the complex plane. For a complex number \( z = a + bi \), the modulus \( |z| \) is found using the formula \( |z| = \sqrt{a^2 + b^2} \). It's essentially the magnitude of the vector representing the complex number.
In the original exercise, the modulus of \( z_1 = 2\sqrt{3} - 2i \) was calculated as \( 4 \), and the modulus of \( z_2 = -1 + i \) was calculated as \( \sqrt{2} \). Knowing the modulus is crucial for expressing complex numbers in polar form and performing operations like multiplication and division.
In the original exercise, the modulus of \( z_1 = 2\sqrt{3} - 2i \) was calculated as \( 4 \), and the modulus of \( z_2 = -1 + i \) was calculated as \( \sqrt{2} \). Knowing the modulus is crucial for expressing complex numbers in polar form and performing operations like multiplication and division.
Argument
The argument of a complex number describes its angle relative to the positive real axis in the complex plane. For a complex number \( z = a + bi \), the argument \( \theta \) can be found using \( \theta = \tan^{-1}(\frac{b}{a}) \).
It's important to consider the quadrant in which the complex number lies, as this affects the sign and value of the argument:
It's important to consider the quadrant in which the complex number lies, as this affects the sign and value of the argument:
- If both \( a \) and \( b \) are positive, \( \theta \) is typically between \( 0 \) and \( \pi/2 \).
- If \( a \) is negative and \( b \) is positive, \( \theta \) lies between \( \pi/2 \) and \( \pi \).
- If both \( a \) and \( b \) are negative, \( \theta \) is typically between \( \pi \) and \( 3\pi/2 \).
- If \( a \) is positive and \( b \) is negative, \( \theta \) lies between \( 3\pi/2 \) and \( 2\pi \).
Product of Complex Numbers
Multiplying complex numbers becomes simpler in polar form. To find the product of two complex numbers \( z_1 \) and \( z_2 \) presented in polar form, we multiply their moduli and add their arguments.
The formula looks like this:
The formula looks like this:
- Modulus of product: \( |z_1z_2| = |z_1| \cdot |z_2| \)
- Argument of product: \( \text{arg}(z_1z_2) = \text{arg}(z_1) + \text{arg}(z_2) \)
Quotient of Complex Numbers
Finding the quotient of two complex numbers is also more straightforward with polar form. The quotient \( \frac{z_1}{z_2} \) is obtained by dividing their moduli and subtracting their arguments.
Here's the breakdown:
Here's the breakdown:
- Modulus of quotient: \( \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \)
- Argument of quotient: \( \text{arg}\left(\frac{z_1}{z_2}\right) = \text{arg}(z_1) - \text{arg}(z_2) \)
