Problem 11 - Analysis 1: Ein Lehr- und Arbeitsbuch für Studienanfänger

Question

Bestimmen sie alle komplexen zahlen z, für die $$ |z+3-3 i|=|z-1-3 i| $$ gilt. Skizzieren Sie die gefundene Punktmenge und beweisen sie Thre Behauptung.

Step-by-Step Solution

Answer

The complex numbers are all on the vertical line \(\text{Re}(z) = -1\).

Step 1: Understand the problem

We are given the condition \[ |z + 3 - 3i| = |z - 1 - 3i| \]and we need to find all complex numbers \( z \) which satisfy this equation. This condition is asking us to find points \( z \) in the complex plane that are equidistant to the points \(-3 + 3i \) and \(1 + 3i \).

Step 2: Interpret geometrically

The equation represents a set of points on the complex plane that are equidistant to two fixed points. Such a set of points forms a perpendicular bisector of the segment connecting those fixed points \(-3 + 3i\) and \(1 + 3i\).

Step 3: Find the midpoint of the segment

The midpoint of the segment connecting \(-3 + 3i\) and \(1 + 3i\) is calculated as follows:\[ \left( \frac{-3+1}{2}, \frac{3+3}{2} \right) = \left( -1, 3 \right). \]

Step 4: Determine the slope of the perpendicular bisector

The line connecting the two points has a horizontal slope since the imaginary parts are equal (both are \(3\)). Therefore, the perpendicular bisector will have a vertical slope. This means the perpendicular bisector is a vertical line.

Step 5: Derive the equation of the line

Since the perpendicular bisector is vertical and passes through the midpoint \(-1 + 3i\), the set of complex numbers \(z = x + yi\) satisfying the equation must have the real part \(x = -1\). Thus, the required line (set of complex numbers \(z\)) is:\[ \text{Re}(z) = -1 \] where \(y\) can be any real number.

Key Concepts

Equidistant Points

In the world of geometry and the complex plane, the term "equidistant points" refers to points that are at the same distance from two fixed positions. In this exercise, we're focused on determining which complex numbers, denoted as \( z \), are equidistant from the fixed points \(-3 + 3i\) and \(1 + 3i\). If you imagine drawing two circles centered at these fixed points, with the same radius, all points \( z \) on the boundary of these circles are equidistant to both centers. However, we're interested here in a particular condition where the radius allows these boundaries to just meet along a line, which leads to the concept of a perpendicular bisector.

Complex Plane

A complex plane, often called the Argand plane, is used to visualize complex numbers. It consists of a horizontal axis (the real axis) and a vertical axis (the imaginary axis). Each complex number \( z = x + yi \) can be represented as a point \( (x, y) \) on this plane.

- **Real part (x):** This is represented on the horizontal axis.
- **Imaginary part (y):** This appears on the vertical axis.

In our exercise, the points \(-3 + 3i\) and \(1 + 3i\) are plotted on the complex plane, reflecting their position dictated by their real and imaginary components. Understanding this layout helps in grasping spatial relationships like distance and collinearity of points, crucial for visualizing how equidistant conditions manifest geometrically.

Perpendicular Bisector

The perpendicular bisector is a fundamental element when finding points equidistant between two locations. In this context, it is the line that is equidistant from both \(-3 + 3i\) and \(1 + 3i\). To construct this line, we start by finding the midpoint of the segment connecting these points.

The midpoint \((-1, 3)\) lies exactly in the middle between the two fixed points along their real components, while sharing the same imaginary component. Once we have the midpoint, the next step is to understand the measured distance runs perpendicular to the line segment's original direction. Because here the line joining the original points is horizontal (same imaginary part), its perpendicular bisector is vertical, signifying completely different orientations.

Midpoint Calculation

Calculating midpoints is essential in geometry to find the center between two positions. In complex numbers, we find it by averaging the real and imaginary components separately. For our given points \(-3 + 3i\) and \(1 + 3i\), the midpoint is calculated as follows:

• **Real part:** \( \frac{-3 + 1}{2} = -1 \)
• **Imaginary part:** \( \frac{3 + 3}{2} = 3 \)

As a result, our midpoint in this scenario is \(-1 + 3i\). This point is crucial since the perpendicular bisector passes through it, helping us to establish the line on which all points equidistant from our given points lie.