Question

Man bestimme alle \(z \in C\) mit $$ z^{3}-i=0 $$

Short answer

The solutions are \( z = \frac{\sqrt{3}}{2} + i\frac{1}{2}, -\frac{\sqrt{3}}{2} + i\frac{1}{2}, -i \).

Step-by-step solution

Recognize the Problem

The problem asks us to find all complex numbers \( z \) such that \( z^3 - i = 0 \). This implies that \( z^3 = i \). This is a root-finding problem for a complex number.

Express the Complex Number in Polar Form

The number \( i \) can be expressed in polar form as \( e^{i(\pi/2 + 2k\pi)} \) where \( k \) is an integer. This accounts for the periodicity of the exponential function in the complex plane.

Apply De Moivre's Theorem

If \( z = re^{i\theta} \) is a complex number in polar form, \( z^3 = r^3 e^{i3\theta} \). So to find \( z \), we set \( r^3 e^{i3\theta} = e^{i(\pi/2 + 2k\pi)} \), implying \( r^3 = 1 \) and \( 3\theta = \pi/2 + 2k\pi \).

Solve for \( r \)

Since \( r^3 = 1 \), \( r \) must be 1, as we are looking for real positive roots (since \( r \) represents magnitude of a complex number).

Solve for \( \theta \)

From \( 3\theta = \pi/2 + 2k\pi \), solve for \( \theta \): \( \theta = (\pi/6) + (2k\pi)/3 \). These are the different possibilities for \( \theta \) when \( k = 0, 1, 2 \) since we need three distinct roots for a cubic equation.

Find the Roots

The different values of \( \theta \) give us the roots: - For \( k = 0 \), \( \theta = \pi/6 \), so \( z_0 = e^{i\pi/6} = \frac{\sqrt{3}}{2} + i\frac{1}{2} \).- For \( k = 1 \), \( \theta = 5\pi/6 \), so \( z_1 = e^{i5\pi/6} = -\frac{\sqrt{3}}{2} + i\frac{1}{2} \).- For \( k = 2 \), \( \theta = 3\pi/2 \), so \( z_2 = e^{i3\pi/2} = -i \).

Key concepts

Polar Form

Complex numbers are extremely useful in many areas of mathematics because they allow us to extend algebraic solutions to equations that have no real solution. The polar form of complex numbers is one of these helpful methods.

Polar form helps us express complex numbers in terms of magnitude and angle, rather than just real and imaginary components. A complex number can be written in the form of \( z = re^{i\theta} \), where \( r \) is the magnitude (or modulus), and \( \theta \) is the angle (or argument) measured in radians from the positive x-axis.

This form is advantageous for multiplication and division of complex numbers because we can simply multiply/divide their magnitudes and add/subtract their angles:

• Magnitude (or modulus) \( r = \sqrt{a^2 + b^2} \)
• Angle (or argument) \( \theta = \arctan\left(\frac{b}{a}\right) \)

For the exercise, we express \( i \), which lies on the imaginary axis, as \( e^{i(\pi/2 + 2k\pi)} \) to account for its periodicity.

De Moivre's Theorem

When it comes to working with complex numbers in polar form, De Moivre's Theorem is your best friend! It simplifies raising a complex number to a power or extracting roots.

De Moivre's Theorem states:\[(z = re^{i\theta})^n = r^n e^{i n\theta}\]This means that to raise a complex number in polar form to the \( n^{th} \) power, you raise its magnitude to the \( n^{th} \) power and multiply its angle by \( n \).
In our exercise, solving the cubic equation \( z^3 = i \) using De Moivre’s Theorem involves equating:- The magnitudes, where we find \( r^3 = 1 \), hence \( r = 1 \).- The angles, leading to \( 3\theta = \pi/2 + 2k\pi \), which gives different \( \theta \) solutions for each \( k \).

This results in three distinct roots, as required by the nature of cubic equations.

Cubic Equation Roots

Finding roots for cubic equations can be tricky, but with complex numbers, it becomes more manageable. In our example exercise, we search for roots of the equation \( z^3 - i = 0 \).

Here's how to find the roots step-by-step:

• First, rewrite the equation as \( z^3 = i \).
• Then, express \( i \) in polar form and apply De Moivre's Theorem to solve for \( z \): \( r^3 e^{i3\theta} = e^{i(\pi/2 + 2k\pi)} \).
• Solve for the magnitude, \( r = 1 \).
• Solve for the angle \( \theta = (\pi/6) + (2k\pi)/3 \) for distinct values of \( k = 0, 1, 2 \).

This method yields three roots, illustrating the fact that a cubic equation has up to three distinct solutions:
- For \( k = 0 \), \( z_0 = e^{i\pi/6} = \frac{\sqrt{3}}{2} + i\frac{1}{2} \).
- For \( k = 1 \), \( z_1 = e^{i5\pi/6} = -\frac{\sqrt{3}}{2} + i\frac{1}{2} \).
- For \( k = 2 \), \( z_2 = e^{i3\pi/2} = -i \).

This comprehensive understanding ensures precise handling of cubic roots in complex numbers.