Problem 6 - Complex Analysis

Question

Sketch the following subsets of \(\mathrm{C}\) in the complex plane: (a) Assume \(a, b \in \mathbb{C}, b \neq 0\); $$ \begin{aligned} &G_{0}:=\left\\{z \in \mathbb{C} ; \quad \operatorname{Im}\left(\frac{z-a}{b}\right)=0\right\\} \\ &G_{+}:=\left\\{z \in \mathbb{C} ; \quad \operatorname{Im}\left(\frac{z-a}{b}\right)>0\right\\} \quad \text { and } \\ &G_{-}:=\left\\{z \in \mathbb{C} ; \quad \operatorname{Im}\left(\frac{z-a}{b}\right)<0\right\\} \end{aligned} $$ (b) Consider \(a, c \in \mathbb{R}\) and \(b \in \mathbb{C}\) with \(b \bar{b}-a c>0\), $$ K:=\\{z \in \mathbb{C} ; \quad a z \bar{z}+\bar{b} z+b \bar{z}+c=0\\} $$ (c) \(L:=\left\\{z \in \mathbb{C} ;\left|z-\frac{\sqrt{2}}{2}\right| \cdot\left|z+\frac{\sqrt{2}}{2}\right|=\frac{1}{2}\right\\}\)

Step-by-Step Solution

Answer

(a) Line and half-planes; (b) Circle; (c) Lemniscate.

Step 1: Identify the Sets for Part (a)

In part (a), we identify the sets based on the imaginary part of the complex expression \(\frac{z-a}{b}\). This is equivalent to finding points for which the imaginary component is 0, positive, or negative. This expression simplifies to \(z = a + bt\), where \(t\) is real. Thus, \(G_0\) is the line through \(a\) in the direction of \(b\); \(G_+\) is the half-plane above this line, and \(G_-\) is the half-plane below it.

Step 2: Sketch the Line and Half-Planes for Part (a)

To sketch these regions in the complex plane, start by drawing the line \(G_0\), defined by \(\operatorname{Im}\left( \frac{z-a}{b} \right) = 0\). This line is parallel to the vector \(b\). Then draw the upper half-plane above this line for \(G_+\), where the imaginary part is positive, and the lower half-plane for \(G_-\), where the imaginary part is negative.

Step 3: Identify and Sketch the Circle in Part (b)

For part (b), the set \(K\) is defined by a quadratic equation. This represents a circle in the complex plane. Use the given condition \(b\bar{b} - ac > 0\) to confirm that the expression represents a circle, as this condition ensures a positive determinant. To sketch \(K\), recognize it as a circle centered at \(-\frac{b}{a}\) with radius determined by adjusting the general circle equation \(x^2 + y^2 = r^2\).

Step 4: Identify the Lemniscate in Part (c)

Part (c) involves a lemniscate, a figure-eight-like curve, defined by \(L = \left\{ z \in \mathbb{C} ; \left| z - \frac{\sqrt{2}}{2} \right| \cdot \left| z + \frac{\sqrt{2}}{2} \right| = \frac{1}{2} \right\}\). Recognize that this expression represents a lemniscate with foci at \( \frac{\sqrt{2}}{2} \) and \( -\frac{\sqrt{2}}{2} \).

Step 5: Sketch the Lemniscate for Part (c)

Draw the lemniscate in the complex plane, centered between \( \frac{\sqrt{2}}{2} \) and \( -\frac{\sqrt{2}}{2} \) along the real axis. The product of distances from any point on the curve to each of these foci equals \( \frac{1}{2} \). This distinctive shape can be outlined using known plots of standard lemniscates.

Key Concepts

Complex Subsets

The concept of complex subsets is fundamental when studying the complex plane. In mathematics, the complex numbers form a plane, known as the complex plane, where each number has a real and an imaginary part.

A complex subset refers to a specific region or subset within this plane. This can be a line, curve, or shape defined by particular properties that all points in the subset share.
For example, in the context of the provided exercise, when we talk about
the subsets \(G_0\), \(G_+\), and \(G_-\), we define each of these in terms of the imaginary part of the transformed complex variable.

• \(G_0\) represents a line where the imaginary part is zero.
• \(G_+\) represents the half-plane where this part is positive.
• \(G_-\) shows the opposite half-plane where the imaginary part is negative.

These subsets provide insights into how complex numbers are distributed and how they interact geometrically in the complex plane.

Imaginary Part

The imaginary part is an essential concept when working with complex numbers. Any complex number \(z\) can be expressed as \(z = x + yi\), where \(x\) is the real part and \(y\) is the imaginary part. The imaginary part determines the vertical position of a point on the complex plane.

Understanding how the imaginary part influences the location of subsets like \(G_0\), \(G_+\), and \(G_-\) can help in visualizing and solving problems involving complex numbers. In the context of the exercise:

• \(\operatorname{Im}(\frac{z-a}{b}) = 0\) indicates that the imaginary part of \({z}\) after transformation is zero.
• This is why \(G_0\) is a straight line, since every point on the line has a neutral imaginary component.
• For \(G_+\), the imaginary part is positive, suggesting a deviation above this line.
• \(G_-\), on the other hand, represents the area where this part is negative, or below the \(G_0\) line.

Visualizing how these imaginary parts split the plane helps students understand the two-dimensional nature of complex numbers.

Quadratic Equation

A quadratic equation in the context of the complex plane takes on a significant geometrical interpretation. Usually expressed in the form \(ax^2 + bx + c = 0\), it can represent various conic sections like circles and ellipses.

For the subset \(K\) in part (b) of the exercise, a quadratic equation represents a circle. This can be described as:

• The equation \(az\bar{z} + \bar{b}z + b\bar{z} + c = 0\) blends real and imaginary terms.

Knowing the conditions for circles helps students understand complex geometry.
With constraints like \(b\bar{b} - ac > 0\), this guides us to confirm that the determinant ensures a circular formation.
The position and size are determined respectively by the values \(a\) and \(c\), with center \(-\frac{b}{a}\). Students should use these parameters to graph the circle intuitively on the complex plane.

Lemniscate

A lemniscate is a fascinating and unique shape that looks like a figure-eight or an infinity symbol.
In mathematics, it is based on the complex plane, where each point satisfies a certain product-distance relationship from two fixed foci.

In the exercise, the lemniscate is outlined by the equation \(L = \left\{ z \in \mathbb{C} ; \left| z - \frac{\sqrt{2}}{2} \right| \cdot \left| z + \frac{\sqrt{2}}{2} \right| = \frac{1}{2} \right\}\). This tells us that any point \(z\) satisfies a particular condition involving the distances to two fixed points \(\frac{\sqrt{2}}{2}\) and \(-\frac{\sqrt{2}}{2}\).

• These are the foci of the lemniscate.
• The product of these distances equals \(\frac{1}{2}\).

Visualizing this peculiar shape and its properties can be complex. However, drawing helps students build intuition around the lemniscate's geometry and its symmetry centered along the real axis in the complex plane.