Question

Quadratwurzeln und Lösbarkeit quadratischer Gleichungen in C Sei \(c=a+i b \neq 0\) eine vorgegebene komplexe Zahl. Durch Aufspaltung in Realund Imaginärteil zeige man, dass es genau zwei verschiedene komplexe Zahlen \({ }_{1} \mathrm{~s}\) und \(z_{2}\) gibt mit $$ z_{1}^{2}=t_{2}^{2}=c . \text { Es ist } z_{2}=-z_{1^{*}} $$ \(\left(z_{1}\right.\) und \(z_{2}\) heiben die Qundratwurzeln aus c.) Als Beispiel bestimme man jeweils die Quadratwurzeln aus. $$ 5+7 i \text { bzw. } \sqrt{2}+i \sqrt{2} $$ Man löse diese Aufgabe auch mit Polarkoordinaten. Ferner zeige man, dass eine quadratische Gleichung $$ x^{2}+\alpha z+\beta=0, \quad \alpha, \beta \in \mathbb{C} \text { beliebig, } $$ stets (h?chstens zwei) Lösungen \(z_{1}, z_{2} \in \mathbb{C}\) besitzt.

Short answer

Each non-zero complex number has two square roots. The quadratic equation solution always exists in \( \mathbb{C} \).

Step-by-step solution

Understanding the Problem

We are given a complex number \( c = a + ib eq 0 \) and need to show that it has exactly two distinct complex roots, \( z_1 \) and \( z_2 \), such that \( z_1^2 = z_2^2 = c \) with \( z_2 = -z_1^* \). Then we'll find square roots of two specific complex numbers and demonstrate using polar coordinates that a quadratic equation has solutions in \( \mathbb{C} \).

General Approach for Complex Numbers

To find square roots of a complex number \( c = a + ib \), we can express \( z = x + yi \) where we need \( x^2 - y^2 + 2xyi = a + ib \). We get the system: \( x^2 - y^2 = a \) and \( 2xy = b \). Solving this system of equations gives us the possible values for \( x \) and \( y \).

Solving for Square Roots

For the two specific numbers, \( 5 + 7i \) and \( \sqrt{2} + i\sqrt{2} \):1. **For \( 5 + 7i \)**: Starting with the system \( x^2 - y^2 = 5 \) and \( 2xy = 7 \), solve these equations. 2. **For \( \sqrt{2} + i\sqrt{2} \)**: Use \( x^2 - y^2 = \sqrt{2} \) and \( 2xy = \sqrt{2} \) to solve the system.

Using Polar Coordinates

Convert the complex numbers to polar form. For example, \( 5+7i \) can be represented as \( r e^{i\theta} \), where \( r = \sqrt{34} \) and \( \theta = \arg(5 + 7i) \). The solutions are \( \sqrt{r} e^{i\theta/2} \) and its conjugate.

Solving Quadratic Equation in \( \mathbb{C} \)

Given the equation \( x^2 + \alpha z + \beta = 0 \), use the quadratic formula for complex numbers: \( z = \frac{-\alpha \pm \sqrt{\alpha^2 - 4\beta}}{2} \). This shows that there can be at most two solutions in \( \mathbb{C} \) depending on the discriminant.

Key concepts

Square Roots of Complex Numbers

When working with complex numbers, finding square roots is an exciting task. A complex number is typically written as \( c = a + ib \), where \( a \) and \( b \) are real numbers. To find its square roots, we are tasked with finding two distinct complex numbers \( z_1 \) and \( z_2 \) such that \( z_1^2 = z_2^2 = c \). This can be done by setting \( z = x + yi \) and solving the system of equations: \( x^2 - y^2 = a \) and \( 2xy = b \). These equations stem from the real and imaginary components of \( z^2 = c \).

Let's consider an example. To find the square roots of \( 5 + 7i \), we set up the system: \( x^2 - y^2 = 5 \) and \( 2xy = 7 \). By solving these simultaneously, we can determine the values of \( x \) and \( y \) that satisfy both conditions. The solutions \( z_1 \) and \( z_2 \) will be conjugates of one another, ensuring the uniqueness of the roots.

Similarly, for \( \sqrt{2} + i\sqrt{2} \), we solve \( x^2 - y^2 = \sqrt{2} \) and \( 2xy = \sqrt{2} \) to find its square roots. Understanding this method allows us to handle any complex number with confidence!

Quadratic Equations in Complex Plane

When dealing with quadratic equations in the complex plane, we're often solving expressions like \( x^2 + \alpha z + \beta = 0 \), where \( \alpha \) and \( \beta \) are complex numbers. The complexity of this problem lies in the fact that the usual real number methods need some adaptation.

In order to find solutions in the complex plane, we apply the quadratic formula: \[ z = \frac{-\alpha \pm \sqrt{\alpha^2 - 4\beta}}{2} \]. Here, the discriminant \( \alpha^2 - 4\beta \) plays a pivotal role. Much like with real numbers, this value will determine whether we have two distinct roots, one double root, or no real roots at all. However, in the complex set \( \mathbb{C} \), a solution always exists, regardless of the discriminant's value.

The novelty in the complex plane is that both solutions are guaranteed because the imaginary unit \( i \) allows the square root of negative numbers to be meaningful. This demonstrates the robustness and elegance of using complex numbers for such quadratic problems.

Polar Coordinates

The concept of polar coordinates offers an insightful way to express complex numbers, facilitating certain calculations, such as finding square roots. When a complex number \( c = a + ib \) is viewed in polar form, it is expressed as \( r e^{i\theta} \), where \( r = \sqrt{a^2 + b^2} \) is the modulus, and \( \theta = \tan^{-1}(\frac{b}{a}) \) is the argument of the complex number.

Polar coordinates are particularly useful because they simplify multiplication and division of complex numbers. For instance, when finding a square root of a complex number, we can use: \( \sqrt{r} e^{i\theta/2} \). The ease of manipulating angles instead of dealing with rectangular coordinates makes complex arithmetic much more intuitive.

Let's illustrate this: given \( 5 + 7i \), converting it to polar form involves finding \( r = \sqrt{34} \) and the angle \( \theta \). The square roots are then given by finding \( \sqrt{r} e^{i\theta/2} \) and its conjugate. By understanding polar coordinates, we open up a new dimension in comprehending and solving complex problems.