Problem 30
Question
Find the modulus \(r\) of the number. Do not use a calculator. $$15 i$$
Step-by-Step Solution
Verified Answer
The modulus is 15.
1Step 1: Understand the problem
We are tasked with finding the modulus \(r\) of the complex number \(15i\).
2Step 2: Recall the formula for modulus
Given a complex number \(a + bi\), its modulus is \(r = \sqrt{a^2 + b^2}\). Here, \(a = 0\) and \(b = 15\) because \(15i\) is equivalent to \(0 + 15i\).
3Step 3: Apply the formula
Substitute \(a = 0\) and \(b = 15\) into the modulus formula: \[ r = \sqrt{0^2 + 15^2} \]
4Step 4: Simplify
Calculate the expression inside the square root: \[ r = \sqrt{0 + 225} = \sqrt{225} \]
5Step 5: Find the square root
Calculate \(\sqrt{225}\), which is \(15\), since \(15^2 = 225\).
Key Concepts
Modulus of Complex NumbersImaginary NumbersSquare Root Calculation
Modulus of Complex Numbers
The modulus of a complex number is a measure of its "size" or absolute value.
This concept is akin to finding the length from the origin to the point representing the complex number in the complex plane.
For complex numbers, represented as \(a + bi\), the modulus \(r\) is calculated by the formula:
In this case, we have the complex number \(15i\), which can be seen as \(0 + 15i\).
This implies:
This concept is akin to finding the length from the origin to the point representing the complex number in the complex plane.
For complex numbers, represented as \(a + bi\), the modulus \(r\) is calculated by the formula:
- \(r = \sqrt{a^2 + b^2}\)
In this case, we have the complex number \(15i\), which can be seen as \(0 + 15i\).
This implies:
- \(a = 0\)
- \(b = 15\)
- \(r = \sqrt{0^2 + 15^2}\)
- \(r = \sqrt{0 + 225}\)
- \(r = \sqrt{225}\)
- \(r = 15\)
Imaginary Numbers
Imaginary numbers are an intriguing concept in mathematics.
They arise when we deal with square roots of negative numbers.
The fundamental imaginary unit is denoted by \(i\), which is defined as \(i = \sqrt{-1}\).
These numbers allow solutions to equations that do not have real solutions.
For example, the imaginary number \(15i\) has no real part (because \(a = 0\)) but represents a unit in the direction of the complex plane.
This visualization helps understand how imaginary numbers extend real numbers to solve new and broader equations.
They arise when we deal with square roots of negative numbers.
The fundamental imaginary unit is denoted by \(i\), which is defined as \(i = \sqrt{-1}\).
- Multiplying \(i\) by itself results in \(i^2 = -1\).
- This means imaginary numbers can express quantities that are otherwise not possible in the realm of real numbers.
These numbers allow solutions to equations that do not have real solutions.
For example, the imaginary number \(15i\) has no real part (because \(a = 0\)) but represents a unit in the direction of the complex plane.
This visualization helps understand how imaginary numbers extend real numbers to solve new and broader equations.
Square Root Calculation
Square root calculation is fundamental in finding the modulus of a complex number and in many other areas of math.
The square root of a number is a value that, when multiplied by itself, gives the original number.
Written as \(\sqrt{x}\), it represents one of the two equal factors of \(x\).
This expertise helps in solving the modulus, among various other problems.
Knowing which squares yield which roots is valuable, especially when calculators aren't an option.
Remembering squares and their roots, like \(15^2 = 225\), provides quick solutions without additional computational tools.
It also deepens understanding of number relationships in mathematical schemas.
The square root of a number is a value that, when multiplied by itself, gives the original number.
Written as \(\sqrt{x}\), it represents one of the two equal factors of \(x\).
- For example, \(\sqrt{225} = 15\) because \(15 \times 15 = 225\).
This expertise helps in solving the modulus, among various other problems.
Knowing which squares yield which roots is valuable, especially when calculators aren't an option.
Remembering squares and their roots, like \(15^2 = 225\), provides quick solutions without additional computational tools.
It also deepens understanding of number relationships in mathematical schemas.
Other exercises in this chapter
Problem 29
Given \(\mathbf{u}=\langle- 2,5\rangle\) and \(\mathbf{v}=\langle 4,3\rangle,\) find each vector. Do not use a calculator. $$v-u$$
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Find a rectangular equation for each curve and graph the curve. $$x=\tan t, y=\cot t ; \text { for } t \text { in }\left(0, \frac{\pi}{2}\right)$$
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Find the cube roots of each complex number. Leave the answers in trigonometric form. Then graph each cube root as a vector in the complex plane. $$\sqrt{3}-i$$
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Solve triangle. \(A=68.41^{\circ}, B=54.23^{\circ}, a=12.75 \mathrm{feet}\)
View solution