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Chapter 9: Problem 2

\(\mathbf{1}-8\) Graph the complex number and find its modulus. $$ -3 i $$

Short Answer

Expert verified
The complex number \(-3i\) is plotted at \((0, -3)\), and its modulus is 3.

Step by step solution

01

Understand the Complex Number

A complex number is expressed in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. Here, the given complex number is \(-3i\), which can be rewritten as \(0 - 3i\). Thus, \(a = 0\) and \(b = -3\).
02

Plot the Complex Number on the Argand Plane

In the Argand plane, the horizontal axis represents the real part and the vertical axis represents the imaginary part. For the complex number \(0 - 3i\), plot the point at the intersection of \(0\) on the real axis and \(-3\) on the imaginary axis.
03

Calculate the Modulus

The modulus of a complex number \(a + bi\) is given by the formula \(\sqrt{a^2 + b^2}\). For the given number \(-3i\), substitute \(a = 0\) and \(b = -3\) into the formula: \(\sqrt{0^2 + (-3)^2} = \sqrt{9} = 3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Argand Plane
The Argand Plane provides a visual representation of complex numbers. Imagine it as a coordinate system where each point represents a complex number.
Here's how it works:
  • The horizontal axis represents the real part of the complex number.
  • The vertical axis represents the imaginary part of the complex number.
In our exercise, the complex number is \(0 - 3i\).
You locate this number on the Argand Plane by finding \(0\) on the real axis and \(-3\) on the imaginary axis.
The beauty of the Argand Plane is that it transforms abstract algebraic concepts into something you can visually comprehend.
If you've ever struggled with understanding complex numbers, plotting them on the Argand Plane might just change that comprehension.
Modulus of a Complex Number
Understanding the modulus of a complex number is key to appreciating the geometry of these numbers.
The modulus, often considered the 'distance' from the origin to the point on the Argand Plane, is found using the formula \(\sqrt{a^2 + b^2}\).
  • For our complex number \(0 - 3i\), we have \(a = 0\) and \(b = -3\).
  • Plug these into the modulus formula: \(\sqrt{0^2 + (-3)^2} = \sqrt{9} = 3\).
This result tells us that the point representing \(0 - 3i\) is exactly 3 units away from the origin on the Argand Plane.
Think of this as the length of the hypotenuse in a right-angled triangle, with both legs rooted in the real and imaginary components!
Imaginary Unit
The imaginary unit, denoted as \(i\), is fundamental to complex numbers.
By definition, \(i\) is such that \(i^2 = -1\). This might seem odd or counterintuitive at first, as we're used to squaring numbers and getting positive results.
The introduction of \(i\) allows for complex numbers like \(a + bi\) to solve equations that real numbers can't, broadening the types of problems we can tackle.
  • In our exercise, \-3i\ is a purely imaginary number, without any real part.
  • It's important to note that even though the number is imaginary, it has a clear representation and behavior within the mathematical framework, thanks to \(i\).
Learning about the imaginary unit enriches your mathematical landscape, offering new ways to represent and interpret numbers.

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Most popular questions from this chapter

Convert the polar equation to rectangular coordinates. $$ r=\frac{4}{1+2 \sin \theta} $$

Convert the polar equation to rectangular coordinates. $$ r=2-\cos \theta $$

(a) Let \(w=\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}\) where \(n\) is a positive integer. Show that \(1, w, w^{2}, w^{3}, \ldots, w^{n-1}\) are the \(n\) distinct \(n\) th roots of \(1 .\) (b) If \(z \neq 0\) is any complex number and \(s^{n}=z,\) show that the \(n\) distinct \(n\) th roots of \(z\) are $$ s, s w, s w^{2}, s w^{3}, \ldots, s w^{n-1} $$

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33-36 Let \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) be vectors and let \(a\) be a scalar. Prove the given property. $$(\mathbf{u}+\mathbf{v}) \cdot \mathbf{w}=\mathbf{u} \cdot \mathbf{w}+\mathbf{v} \cdot \mathbf{w}$$
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