Problem 7 - Complex Analysis

Question

Let \(c=a+\mathrm{i} b \neq 0\) be a given complex number. By splitting it into its real and imaginary parts show that there are exactly two complex numbers \(z_{1}\) and \(z_{2}\) such that $$ z_{1}^{2}=z_{2}^{2}=c . \text { We have } z_{2}=-z_{1} $$ \(\left(z_{1}\right.\) and \(z_{2}\) are called the square roots of \(\left.c .\right)\) For example, determine the square roots of $$ 5+7 \mathrm{i}, \quad \text { and } \quad \sqrt{2}+\mathrm{i} \sqrt{2} $$ Use polar coordinates for this exercise. Furthermore, show that a quadratic equation $$ z^{2}+\alpha z+\beta=0, \quad \alpha, \beta \in \mathbb{C} \text { arbitrary } $$ always has at most two solutions \(z_{1}, z_{2} \in \mathbb{C}\).

Step-by-Step Solution

Answer

The square roots of a complex number and roots of a quadratic equation show there are at most two complex solutions.

Step 1: Write the Complex Number in Polar Form

Let the complex number be \( c = a + bi \). In polar form, a complex number is represented as \( c = r(\cos\theta + i\sin\theta) \) where \( r = \sqrt{a^2 + b^2} \) is the modulus and \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \) is the argument of \( c \).

Step 2: Find Square Roots Using De Moivre's Theorem

Using De Moivre's Theorem, the square roots \( z_1 \) and \( z_2 \) of \( c \) are given by \( z_k = \sqrt{r}(\cos(\frac{\theta + 2k\pi}{2}) + i\sin(\frac{\theta + 2k\pi}{2})) \) for \( k = 0, 1 \).

Step 3: Verify the Condition \(z_1^2 = z_2^2 = c\)

Since \( z_1 = \sqrt{r}(\cos(\frac{\theta}{2}) + i\sin(\frac{\theta}{2})) \) and \( z_2 = \sqrt{r}(\cos(\frac{\theta + \pi}{2}) + i\sin(\frac{\theta + \pi}{2})) = -z_1 \), it follows that \( z_1^2 = (\sqrt{r})^2(\cos\theta + i\sin\theta) = c \) and \( z_2^2 = (-z_1)^2 = z_1^2 = c \), verifying the statement given in the problem.

Step 4: Calculate Example Square Roots

For \( c = 5 + 7i \), calculate \( r = \sqrt{5^2 + 7^2} = \sqrt{74} \) and \( \theta = \tan^{-1}(\frac{7}{5}) \). Then calculate \( z_1 \) and \( z_2 = -z_1 \). Similarly, for \( c = \sqrt{2} + i\sqrt{2} \), calculate \( r \) and \( \theta \), then \( z_1 \) and \( z_2 = -z_1 \).

Step 5: Formulate and Solve the Quadratic Equation

Consider the quadratic equation \( z^2 + \alpha z + \beta = 0 \). The solutions are given by the quadratic formula: \( z = \frac{-\alpha \pm \sqrt{\alpha^2 - 4\beta}}{2} \). Since this formula produces at most two distinct solutions, it shows that the equation always has at most two roots in \( \mathbb{C} \).

Key Concepts

Polar Coordinates

When dealing with complex numbers, converting them into polar coordinates can simplify calculations. A complex number in the form of \(a + bi\) can be represented in polar form as \(r(\cos\theta + i\sin\theta)\). Here, \(r\) is known as the modulus and is calculated as \(\sqrt{a^2 + b^2}\).
\(\theta\), the argument, is computed using \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\). This transformation helps with the rotation of complex numbers on a plane, aligning them based on their distance from the origin and their angle relative to the positive x-axis.
Using polar coordinates is especially useful in operations like finding powers and roots of complex numbers, as it makes their mathematical manipulation more straightforward.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool in the realm of complex numbers, allowing us to efficiently compute powers and roots. It states that for a complex number in polar form, \(z = r(\cos\theta + i\sin\theta)\), the \(n^{th}\) power of \(z\) is given by \[z^n = r^n (\cos(n\theta) + i\sin(n\theta))\].
This theorem significantly aids in calculating roots, like square roots, by setting \(n = 1/2\). This process involves simplifying expressions since multiplying angles and moduli becomes straightforward while maintaining their trigonometric relationships.

Quadratic Equations

A quadratic equation in the complex number context extends the standard form \(z^2 + \alpha z + \beta = 0\). The solutions can be found using the quadratic formula: \[z = \frac{-\alpha \pm \sqrt{\alpha^2 - 4\beta}}{2}\]. This equation can yield two solutions, revealing the roots of the function.
Complex quadratic equations, like real ones, may have distinct, repeated, or even complex conjugate roots, highlighting their versatility but also the complexity in considering factors such as complex discriminants.