Question

Solve each of the following equations leaving (a) \(z^{3}=-1\) (b) \(z^{3}=1\) (c) \(z^{3}-6 \mathrm{j}=0\) your answers in polar form: (a) \(z^{2}=1+\mathrm{j}\) (b) \(z^{2}=1-\mathrm{j}\) (c) \(z^{3}=-2+3 \mathrm{j}\)

Short answer

(a) The cube roots of -1 are: $$z_1=\sqrt[3]{1}(\cos(\frac{\pi}{3})+j\sin(\frac{\pi}{3}))$$ $$z_2=\sqrt[3]{1}(\cos(\frac{3\pi}{3})+j\sin(\frac{3\pi}{3}))$$ $$z_3=\sqrt[3]{1}(\cos(\frac{5\pi}{3})+j\sin(\frac{5\pi}{3}))$$ (b) The cube roots of 1 are: $$z_1=\sqrt[3]{1}(\cos(\frac{0}{3})+j\sin(\frac{0}{3}))$$ $$z_2=\sqrt[3]{1}(\cos(\frac{2\pi}{3})+j\sin(\frac{2\pi}{3}))$$ $$z_3=\sqrt[3]{1}(\cos(\frac{4\pi}{3})+j\sin(\frac{4\pi}{3}))$$ (c) The cube roots of 6j are: $$z_1=\sqrt[3]{6}(\cos(\frac{\pi}{6})+j\sin(\frac{\pi}{6}))$$ $$z_2=\sqrt[3]{6}(\cos(\frac{5\pi}{6})+j\sin(\frac{5\pi}{6}))$$ $$z_3=\sqrt[3]{6}(\cos(\frac{9\pi}{6})+j\sin(\frac{9\pi}{6}))$$

Step-by-step solution

Convert -1 to polar form

To find the polar form of -1, we can write it as \(-1+0j\). The magnitude is 1 and the angle is \(\pi\). So, \(-1 =1(\cos\pi + j\sin\pi)\). #Step 2: Apply De Moivre's Theorem#

Find the cube roots of -1

To find the cube roots, we will use De Moivre's Theorem: $$z=\sqrt[3]{r}(\cos(\frac{\theta+2k\pi}{n})+j\sin(\frac{\theta+2k\pi}{n}))$$ Here, \(n=3\), \(r=1\), and \(\theta=\pi\). Substituting the values and plugging in \(k=0,1,2\) gives us three cube roots: $$z_1=\sqrt[3]{1}(\cos(\frac{\pi}{3})+j\sin(\frac{\pi}{3}))$$ $$z_2=\sqrt[3]{1}(\cos(\frac{3\pi}{3})+j\sin(\frac{3\pi}{3}))$$ $$z_3=\sqrt[3]{1}(\cos(\frac{5\pi}{3})+j\sin(\frac{5\pi}{3}))$$ #b. Solving for z^3= 1# #Step 1: Convert to polar form#

Convert 1 to polar form

To find the polar form of 1, we can write it as \(1+0j\). The magnitude is 1 and the angle is \(0\). So, \(1 =1(\cos0 + j\sin0)\). #Step 2: Apply De Moivre's Theorem#

Find the cube roots of 1

Using De Moivre's Theorem with \(n=3\), \(r=1\), and \(\theta=0\), we get three cube roots: $$z_1=\sqrt[3]{1}(\cos(\frac{0}{3})+j\sin(\frac{0}{3}))$$ $$z_2=\sqrt[3]{1}(\cos(\frac{2\pi}{3})+j\sin(\frac{2\pi}{3}))$$ $$z_3=\sqrt[3]{1}(\cos(\frac{4\pi}{3})+j\sin(\frac{4\pi}{3}))$$ #c. Solving for z^3 - 6j= 0# #Step 1: Convert to polar form#

Convert 6j to polar form

To find the polar form of 6j, we can write it as \(0+6j\). The magnitude is 6 and the angle is \(\frac{\pi}{2}\). So, \(6j =6(\cos(\frac{\pi}{2}) + j\sin(\frac{\pi}{2}))\). #Step 2: Apply De Moivre's Theorem#

Find the cube roots of 6j

Using De Moivre's Theorem with \(n=3\), \(r=6\), and \(\theta=\frac{\pi}{2}\), we get three cube roots: $$z_1=\sqrt[3]{6}(\cos(\frac{\pi}{6})+j\sin(\frac{\pi}{6}))$$ $$z_2=\sqrt[3]{6}(\cos(\frac{5\pi}{6})+j\sin(\frac{5\pi}{6}))$$ $$z_3=\sqrt[3]{6}(\cos(\frac{9\pi}{6})+j\sin(\frac{9\pi}{6}))$$ Note: Square roots for three complex numbers can be computed in a similar way, by finding the necessary polar forms and using De Moivre's Theorem with \(n=2\).

Key concepts

Polar Form

Polar form is a way of expressing complex numbers, allowing for easy multiplication and division. Complex numbers have a real part and an imaginary part, like \(a + bj\), where \(a\) is the real part and \(b\) is the imaginary part. The polar form expresses these numbers in terms of magnitude and angle. To convert a complex number into polar form, you must find two things:

• Magnitude: The magnitude (or modulus) is the distance of the number from the origin in the complex plane. It is calculated as \( \sqrt{a^2 + b^2} \).
• Argument: The angle (or argument) from the positive real axis to the line representing the complex number is determined using the inverse tangent function: \( \theta = \tan^{-1}(\frac{b}{a}) \).

Once you have both magnitude and argument, the polar form can be expressed as \( r (\cos \theta + j \sin \theta) \,\) where \(r\) is the magnitude and \(\theta\) is the angle. This form is especially useful while performing powers and roots of complex numbers, thanks to De Moivre's Theorem.

When solving problems, ensure to take care of the angle's quadrant by checking the sign of parts \(a\) and \(b\). Remember, real numbers are also complex numbers with an imaginary part of zero (e.g., \( -1 + 0j \) in polar form becomes \(1 (\cos \pi + j \sin \pi) \)).

De Moivre's Theorem

De Moivre's Theorem is a powerful formula used to find powers and roots of complex numbers in polar form. This theorem states that for a complex number expressed as \( r(\cos \theta + j \sin \theta) \), the \(n\)-th power of the complex number can be expressed as:

\[r^n(\cos(n\theta) + j \sin(n\theta))\]

Similarly, to find the \(n\)-th roots of a complex number, the theorem is extended to:

\[ \sqrt[n]{r}(\cos(\frac{\theta + 2k\pi}{n}) + j \sin(\frac{\theta + 2k\pi}{n}))\]

where \(k = 0, 1, 2, ..., n-1\). This generates all possible \(n\)-th roots of the number.

In the context of the exercise, De Moivre’s theorem is applied to solve \((z^3 = -1)\) by first converting \(-1\) to polar form, then using the theorem to find three distinct cube roots. This method is systematic and greatly simplifies complex equations of powers like finding cube roots, allowing you to solve without excessive algebraic manipulation.

Cube Roots

When dealing with complex equations, finding cube roots is a common task. A cube root of a number \(z\) is any complex number that, when raised to the third power, yields \(z\). Using the polar form along with De Moivre's Theorem ensures a straightforward path to finding all cube roots of a complex number.

Consider a complex number \(z\) in polar form \(r (\cos \theta + j \sin \theta)\). To find the cube roots, apply the formula:

• Begin with the cube root of the magnitude: \(\sqrt[3]{r}\)
• Calculate the angle for \(k = 0, 1, 2\) using: \(\frac{\theta + 2k\pi}{3}\)

This process results in three roots that are symmetrically spaced about the origin in the complex plane. For example, finding the cube roots of \(-1\) involves using the polar angle of \(\pi\) to calculate:

\{(\cos(\frac{\pi}{3})+j\sin(\frac{\pi}{3}), \cos(\frac{\pi}{3}+2\frac{\pi}{3})+j\sin(\frac{\pi}{3}+2\frac{\pi}{3}), \cos(\frac{\pi}{3}+4\frac{\pi}{3})+j\sin(\frac{\pi}{3}+4\frac{\pi}{3}) )\}.

Understanding this symmetry helps to visualize the roots' locations in the complex plane and to confirm that these solutions are indeed valid cube roots.

Complex Equations

Solving complex equations often involves more than just basic algebraic manipulation; you need a good understanding of complex numbers and their representations. With complex equations involving powers, like \(z^3 = 1\) or \(z^3 = -2 + 3j\), it's crucial to employ techniques like De Moivre's Theorem to simplify the process.

To tackle such equations:

• Convert the given complex number into polar form, identifying the magnitude and angle.
• Apply De Moivre’s Theorem to calculate the required power or root of the number.
• Find all possible solutions by considering different values of \(k\).

These equations often have multiple solutions due to the periodic nature of trigonometric functions. Using polar forms helps in easily identifying these solutions. For instance, the task of finding cube roots involves calculating multiple angles for potential solutions. Each angle and corresponding root represents one of the solutions to the complex equation.

Be mindful of the polar angles' periodicity, aiding in understanding the full set of roots. This comprehensiveness allows for solving not just simple expressions but more intricate ones, like those with combinations of real and imaginary parts.