Question

Describe geometrically the set of points in the complex plane satisfying the following equations. $$|z-1|<1$$

Short answer

Interior of a circle centered at (1,0) with radius 1.

Step-by-step solution

Identify the complex number components

Recall that a complex number can be written as $z=x+yi$, where $x$ is the real part and $y$ is the imaginary part.

Interpret the given equation geometrically

The given equation is $$|z-1|<1$$. This can be rewritten to $$|(x+yi)-1|<1$$

Simplify the expression inside the magnitude

Rewrite $(x+yi)-1$ as $(x-1)+yi$, so the equation becomes $$|(x-1)+yi|<1$$

Understand magnitude in terms of distance

The magnitude $$|(x-1)+yi|$$ represents the distance from the point $(x,y)$ to the point $(1,0)$ in the complex plane.

Describe the geometrical shape

The inequality $$|(x-1)+yi|<1$$ translates to all points $(x,y)$ whose distance from $(1,0)$ is less than 1. This describes the interior of a circle centered at $(1,0)$ with radius 1.

Key concepts

Complex Numbers

A complex number is a mathematical expression of the form $z=x+yi$, where $x$ is the real part and $y$ is the imaginary part of the number. Complex numbers are often represented in the complex plane, where the horizontal axis denotes the real part and the vertical axis denotes the imaginary part.
This makes visualizing and analyzing complex numbers much easier. The shorthand notation for the imaginary unit is $i$, and it's defined as $i^2=-1$.
Every complex number can be uniquely located in the complex plane based on its real and imaginary components.

Inequalities

An inequality in the context of complex numbers often means specifying a set of points that satisfy a certain condition. In this exercise, the inequality is $|z-1|<1$. This notation means we are looking at all complex numbers $z$ such that their magnitude after subtracting a specific number is less than 1.
In simpler terms, the magnitude $|z-1|$ represents the distance between the complex number $z$ and the fixed number 1 (which can be represented as $1+0i$) in the complex plane.

Distance in the Complex Plane

The distance between two points in the complex plane can be calculated using the concept of magnitude. Given two complex numbers $z_1=x_1+y_{1}i$ and $z_2=x_2+y_{2}i$, their distance is $|z_1-z_2|$. This is analogous to the Euclidean distance formula in a coordinate system.
The magnitude of a complex number $a+bi$ is given by $\sqrt{a^2+b^2}$. For this exercise, the expression $|z-1|<1$ represents the set of all points whose distance from $1$ is less than 1 unit. This concept is a crucial bridge connecting algebraic inequalities with geometric interpretations.

Geometric Interpretation

Geometrically, the inequality $|z-1|<1$ represents a circle centered at the point $(1,0)$ with radius 1 in the complex plane. This is because $|z-1|$ translates to the distance from $z$ to $1$, and the inequality states that this distance must be less than 1.
So, when we say $|z-1|<1$, we are referring to all the points inside a circle of radius 1, centered at $1$ on the real axis. To visualize, imagine drawing a circle on a grid: all points within that circle (not including the boundary) satisfy the given inequality.