Finding Points In The Complex Plane: A Step-by-Step Guide

Hey math enthusiasts! Today, we're diving into the world of complex numbers and exploring how to determine the sets of points in the complex plane based on given conditions. Specifically, we'll be tackling problems related to the modulus (or absolute value) of a complex number, denoted as |z|. This guide is designed to break down each scenario, making it easy for you to visualize and understand the solutions. Let's jump right in and make complex numbers a breeze!

Understanding the Modulus of a Complex Number

Before we start, let's quickly recap what the modulus of a complex number means. If we have a complex number z = x + yi, where x is the real part and y is the imaginary part, the modulus |z| is the distance of the point (x, y) from the origin (0, 0) in the complex plane. It's calculated as |z| = √(x² + y²). Think of it like the radius of a circle centered at the origin. This understanding is super important because the conditions we'll be dealing with, such as |z| = 3, |z| ≤ 4, etc., directly relate to this distance.

Now, let's go through each part of the problem step by step to determine the sets of points. I will show you how to solve them. These steps will make you a pro at solving these problems. Let's do it!

a) |z| = 3: The Circle with Radius 3

In this case, the condition |z| = 3 tells us that the distance of the point z from the origin is exactly 3. Imagine all the points that are precisely 3 units away from the origin. What shape do they form? That's right, it's a circle! So, the set of points that satisfy this condition is a circle centered at the origin (0, 0) with a radius of 3. All complex numbers z that lie on this circle meet the condition. If z = x + yi, then √(x² + y²) = 3, and squaring both sides, we get x² + y² = 9. This is the equation of the circle.

Therefore, the set of points is a circle centered at the origin with a radius of 3. Easy, peasy, right? This problem is great because you can visualize the result easily. You will master all of these problems in no time! This understanding is great because it connects algebra with geometry in a really visual way. We're not just crunching numbers; we're seeing shapes emerge! This approach allows you to solve different exercises because you will easily see what they are asking.

b) |z| ≤ 4: The Disk with Radius 4

Now, let's spice things up a bit! The condition is now |z| ≤ 4. This means the distance of z from the origin is less than or equal to 4. This includes all points that are 4 units away (which is our circle from the previous example) and all points inside that circle. So, we're not just looking at a circle's perimeter; we're also including its interior. This set of points is a closed disk (a filled-in circle) centered at the origin with a radius of 4. The inequality means that the radius can be less than or equal to 4. If z = x + yi, then √(x² + y²) ≤ 4, and squaring both sides, we get x² + y² ≤ 16. This inequality represents the disk.

So, the set of points is a closed disk centered at the origin with a radius of 4. Think of it as drawing a circle of radius 4 and then shading in the entire area inside the circle, including the circumference. This is super useful because it allows you to visualize how the modulus relates to areas in the complex plane, not just lines. You will do great in the exam!

c) |z| ≥ 2: The Exterior of a Circle with Radius 2

Alright, let's shift gears again. This time, we have |z| ≥ 2. This means the distance of z from the origin is greater than or equal to 2. It encompasses all the points on a circle of radius 2 and all the points outside that circle. So, we have a circle of radius 2, and then we extend our area outwards, away from the origin, covering all points farther away. If z = x + yi, then √(x² + y²) ≥ 2, and squaring both sides, we get x² + y² ≥ 4. This inequality represents the exterior of the circle, including the circle itself.

Consequently, the set of points is the exterior of a circle centered at the origin with a radius of 2, including the circle itself. Imagine drawing a circle of radius 2 and then shading everything outside of it. The points on the circle are also part of the solution because of the “equal to” part of the inequality. This exercise teaches you that not only the interior is important, but also the exterior part of the circles is important. Keep that in mind because it is very important for complex number problems.

d) 1 ≤ |z|: The Annulus (Ring)

Here comes the fun part! The condition is now 1 ≤ |z|. This means the distance of z from the origin is greater than or equal to 1. This actually combines two ideas: it includes all points that are at least a distance of 1 from the origin. The most challenging part about this is that it's a combination of previous problems. This condition defines a region that is everything outside a circle of radius 1, including the circle, and also inside a