Chapter 7: Problem 43
Write each complex number in the form \(a+b i\). $$\sqrt{6}\left(\cos 60^{\circ}+i \sin 60^{\circ}\right)$$
Short Answer
Expert verified
\( \frac{\sqrt{6}}{2} + i \frac{3\sqrt{2}}{2} \)
Step by step solution
01
- Identify the given complex number in polar form.
The complex number is given in polar form as \ \( \sqrt{6}(\cos 60^{\circ} + i \sin 60^{\circ}) \). Here, \( r = \sqrt{6} \), \( \theta = 60^{\circ} \), and it can be expressed as \( r(\cos \theta + i \sin \theta) \).
02
- Convert the trigonometric functions to their respective values.
Using the known values for cosine and sine at 60 degrees, \ \( \cos 60^{\circ} = \frac{1}{2} \) and \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \).
03
- Substitute the trigonometric values into the expression.
Substitute \( \cos 60^{\circ} \) and \( \sin 60^{\circ} \) into the given expression: \ \( \sqrt{6}(\frac{1}{2} + i \frac{\sqrt{3}}{2}) \).
04
- Distribute \( \sqrt{6} \) to both parts inside the parenthesis.
Multiply \( \sqrt{6} \) by each part of the expression: \ \( \sqrt{6} \cdot \frac{1}{2} + \sqrt{6} \cdot i \frac{\sqrt{3}}{2} \).
05
- Simplify the expression.
Perform the multiplication: \ \( \sqrt{6} \cdot \frac{1}{2} = \frac{\sqrt{6}}{2} \) and \( \sqrt{6} \cdot i \frac{\sqrt{3}}{2} = i \frac{\sqrt{18}}{2} = i\frac{3\sqrt{2}}{2} \). So, the simplified expression is \( \frac{\sqrt{6}}{2} + i \frac{3\sqrt{2}}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Form
Polar form is a way to express complex numbers which are numbers in the form of \( a + bi \). This form is particularly useful in multiplication and division of complex numbers because it leverages the magnitude (or modulus) and the angle of the complex number.
In polar form, a complex number is represented as \( r(\text{cos} \theta + i \text{sin} \theta) \), where \( r \) is the magnitude and \( \theta \) is the argument or angle.
For example, given the complex number \( \sqrt{6} (\text{cos} 60^{\circ} + i \text{sin} 60^{\circ}) \), we see that:
This establishes an easily recognizable structure for complex numbers, making operations more intuitive. It’s especially useful for converting into a more standard form, like \( a + bi \).
In polar form, a complex number is represented as \( r(\text{cos} \theta + i \text{sin} \theta) \), where \( r \) is the magnitude and \( \theta \) is the argument or angle.
For example, given the complex number \( \sqrt{6} (\text{cos} 60^{\circ} + i \text{sin} 60^{\circ}) \), we see that:
- \( r = \sqrt{6} \)
- \( \theta = 60^{\circ} \)
This establishes an easily recognizable structure for complex numbers, making operations more intuitive. It’s especially useful for converting into a more standard form, like \( a + bi \).
Trigonometric Conversion
Trigonometric conversion is the process of using trigonometric function values to convert from polar form to rectangular form (\( a + bi \)). In our example:
By substituting these values back into the polar form expression, we have:
\( \sqrt{6} (\text{cos} 60^{\circ} + i \text{sin} 60^{\circ}) = \sqrt{6} (\frac{1}{2} + i \frac{\sqrt{3}}{2}) \)
This step makes the equation more manageable by transforming it from an abstract form involving trigonometric functions into a more familiar algebraic expression. This is a key part of solving and understanding the complex number.
- \( \text{cos} 60^{\circ} = \frac{1}{2} \)
- \( \text{sin} 60^{\circ} = \frac{\sqrt{3}}{2} \)
By substituting these values back into the polar form expression, we have:
\( \sqrt{6} (\text{cos} 60^{\circ} + i \text{sin} 60^{\circ}) = \sqrt{6} (\frac{1}{2} + i \frac{\sqrt{3}}{2}) \)
This step makes the equation more manageable by transforming it from an abstract form involving trigonometric functions into a more familiar algebraic expression. This is a key part of solving and understanding the complex number.
Simplification
To fully simplify a complex equation, you need to break down and combine the terms efficiently. In our given example, once we substitute with trigonometric values, we simplify it further by distributing \( \sqrt{6} \):
\( \sqrt{6} (\frac{1}{2} + i \frac{\sqrt{3}}{2}) \)
Distribute \( \sqrt{6} \):
\( \sqrt{6} \cdot \frac{1}{2} + \sqrt{6} \cdot i \frac{\sqrt{3}}{2} \)
Performing the multiplication gives:
So, it simplifies to:
\( \frac{\sqrt{6}}{2} + i \frac{3\sqrt{2}}{2} \).
This final form is much simpler and is expressed in the standard format of complex numbers, \( a + bi \). Understanding the importance of these steps - from substitution to distribution, and then the final multiplication - is vital in mastering the simplification of complex numbers.
\( \sqrt{6} (\frac{1}{2} + i \frac{\sqrt{3}}{2}) \)
Distribute \( \sqrt{6} \):
\( \sqrt{6} \cdot \frac{1}{2} + \sqrt{6} \cdot i \frac{\sqrt{3}}{2} \)
Performing the multiplication gives:
- \( \sqrt{6} \cdot \frac{1}{2} = \frac{\sqrt{6}}{2} \)
- \( \sqrt{6} \cdot i \frac{\sqrt{3}}{2} = i \frac{\sqrt{18}}{2} = i \frac{3\sqrt{2}}{2} \)
So, it simplifies to:
\( \frac{\sqrt{6}}{2} + i \frac{3\sqrt{2}}{2} \).
This final form is much simpler and is expressed in the standard format of complex numbers, \( a + bi \). Understanding the importance of these steps - from substitution to distribution, and then the final multiplication - is vital in mastering the simplification of complex numbers.