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Magazine Chapter 10: Problem 25
Write \(z_{1}\) and \(z_{2}\) in polar form, and then find the reciprocals \(\frac{1}{z_{1}}, \frac{1}{z_{2}^{\prime}}\) the product \(z_{1} z_{2}\) and the quotient \(\frac{z_{1}}{z_{2}}(-\pi<\theta<\pi)\). $$z_{1}=\sqrt{3}+i \text { and } z_{2}=2-2 i \sqrt{3}$$
Short Answer
Expert verified
\(\frac{1}{z_1} = \frac{\sqrt{3}}{4} - \frac{i}{4}\), \(\frac{1}{z_2} = \frac{1}{8} + \frac{i\sqrt{3}}{8}\); \(z_1 z_2 = \sqrt{3} - i\); \(\frac{z_1}{z_2} = \frac{i}{2}\).
Step by step solution
01
Convert to Polar Form
To express the complex numbers in polar form, we need the modulus and argument for each: For \(z_1 = \sqrt{3} + i\):- Modulus: \(r_1 = |z_1| = \sqrt{(\sqrt{3})^2 + 1^2} = 2\).- Argument: \(\theta_1 = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\) (since both components are positive).For \(z_2 = 2 - 2i\sqrt{3}\):- Modulus: \(r_2 = |z_2| = \sqrt{2^2 + (-2\sqrt{3})^2} = 4\).- Argument: \(\theta_2 = \tan^{-1}\left(\frac{-2\sqrt{3}}{2}\right) = -\frac{\pi}{3}\).So, in polar form:\[ z_1 = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) \]\[ z_2 = 4\left(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})\right) \]
02
Find Reciprocals
To find the reciprocals, use the polar form:For \(z_1\): - Reciprocal: \(\frac{1}{z_1} = \frac{1}{2}\left(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})\right)\), which simplifies to \(\frac{\sqrt{3}}{4} - \frac{i}{4}\).For \(z_2\):- Reciprocal: \(\frac{1}{z_2} = \frac{1}{4}\left(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})\right)\), which simplifies to \(\frac{1}{8} + \frac{i\sqrt{3}}{8}\).
03
Calculate the Product
The product of two complex numbers in polar form is the product of their moduli and the sum of their arguments:- Modulus: \(2 \times 4 = 8\)- Argument: \(\frac{\pi}{6} + (-\frac{\pi}{3}) = -\frac{\pi}{6}\)Thus, the product is:\[ z_1 z_2 = 8\left(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})\right) \]Simplifying, we have:\[ z_1 z_2 = \sqrt{3} - i \]
04
Calculate the Quotient
The quotient of two complex numbers in polar form involves dividing the moduli and subtracting the arguments:- Modulus: \(\frac{2}{4} = \frac{1}{2}\)- Argument: \(\frac{\pi}{6} - (-\frac{\pi}{3}) = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}\)So, the quotient is:\[ \frac{z_1}{z_2} = \frac{1}{2}\left(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})\right) \]This simplifies to:\[ \frac{z_1}{z_2} = \frac{i}{2} \]
05
Summary of Results
- Reciprocals: \(\frac{1}{z_1} = \frac{\sqrt{3}}{4} - \frac{i}{4}\), \(\frac{1}{z_2} = \frac{1}{8} + \frac{i\sqrt{3}}{8}\).- Product: \(z_1 z_2 = \sqrt{3} - i\).- Quotient: \(\frac{z_1}{z_2} = \frac{i}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Form
Complex numbers can be expressed in both Cartesian form, such as \( a + bi \), and polar form. The polar form is particularly useful in multiplication, division, and finding powers or roots of complex numbers.
In polar form, a complex number \( z = x + yi \) is represented as \( z = r \left( \cos \theta + i \sin \theta \right) \). This representation highlights:
In polar form, a complex number \( z = x + yi \) is represented as \( z = r \left( \cos \theta + i \sin \theta \right) \). This representation highlights:
- \( r \), the modulus of \( z \), which is the distance from the origin to the point \((x, y)\) in the complex plane.
- \( \theta \), the argument of \( z \), which is the angle the line makes with the positive x-axis.
Modulus
The modulus of a complex number \( z = x + yi \) is a measure of its magnitude. It is calculated using the formula:
\[ r = |z| = \sqrt{x^2 + y^2} \]
The modulus represents the distance from the origin to the point \((x, y)\) in the complex plane. The formula resembles the Pythagorean theorem and gives a non-negative real number.
\[ r = |z| = \sqrt{x^2 + y^2} \]
The modulus represents the distance from the origin to the point \((x, y)\) in the complex plane. The formula resembles the Pythagorean theorem and gives a non-negative real number.
- For \( z_1 = \sqrt{3} + i \), the modulus is \( 2 \).
- For \( z_2 = 2 - 2i\sqrt{3} \), the modulus is \( 4 \).
Argument
The argument of a complex number \( z = x + yi \) is the angle \( \theta \) formed with the positive x-axis in the complex plane. The argument is determined using the inverse tangent function:
\[ \theta = \tan^{-1} \left( \frac{y}{x} \right) \]
This calculation provides the angle in radians, and it can be either positive or negative depending on the quadrant.
\[ \theta = \tan^{-1} \left( \frac{y}{x} \right) \]
This calculation provides the angle in radians, and it can be either positive or negative depending on the quadrant.
- The argument for \( z_1 = \sqrt{3} + i \) is \( \frac{\pi}{6} \).
- The argument for \( z_2 = 2 - 2i\sqrt{3} \) is \( -\frac{\pi}{3} \).
Reciprocal
The reciprocal of a complex number \( z \) in polar form is given by:
\[ \frac{1}{z} = \frac{1}{r} \left( \cos(-\theta) + i \sin(-\theta) \right) \]
This formula leverages the properties of polar form for simplification. The reciprocal's modulus is the inverse of the original modulus, and the argument becomes its negative.
\[ \frac{1}{z} = \frac{1}{r} \left( \cos(-\theta) + i \sin(-\theta) \right) \]
This formula leverages the properties of polar form for simplification. The reciprocal's modulus is the inverse of the original modulus, and the argument becomes its negative.
- The reciprocal of \( z_1 \) in polar form becomes \( \frac{\sqrt{3}}{4} - \frac{i}{4} \).
- The reciprocal of \( z_2 \) simplifies to \( \frac{1}{8} + \frac{i\sqrt{3}}{8} \).
Product and Quotient of Complex Numbers
Multiplying and dividing complex numbers become easier in polar form. For the product \( z_1 \times z_2 \):
For this exercise:
- Multiply their moduli: \( r_1 \times r_2 \).
- Add their arguments: \( \theta_1 + \theta_2 \).
- Divide their moduli: \( \frac{r_1}{r_2} \).
- Subtract their arguments: \( \theta_1 - \theta_2 \).
For this exercise:
- The product \( z_1 z_2 \) is \( \sqrt{3} - i \) with modulus \( 8 \) and argument \( -\frac{\pi}{6} \).
- The quotient \( \frac{z_1}{z_2} \) is \( \frac{i}{2} \) with modulus \( \frac{1}{2} \) and argument \( \frac{\pi}{2} \).
