Powers of Complex Numbers (DeMoivre's Theorem): Videos & Practice Problems

Complex numbers in polar form can be efficiently manipulated using de Moivre's theorem, which states that raising a complex number to a power involves raising the modulus to that power and multiplying the angle by the same power. For roots, multiple solutions arise, calculated as $r^{\frac{1}{n}}cis(\frac{1}{n}\theta +360k)$, where k ranges from 0 to n-1, providing all possible roots.

Powers Of Complex Numbers In Polar Form (DeMoivre's Theorem)

Complex numbers can be expressed in polar form, which is particularly useful for performing operations like multiplication and exponentiation. When multiplying complex numbers in polar form, the process involves taking the product of their magnitudes (r values) and adding their angles. However, a more efficient method exists for raising complex numbers to a power, known as de Moivre's theorem.

According to de Moivre's theorem, to raise a complex number in polar form to a power \( n \), you follow these steps: first, raise the magnitude \( r \) to the power \( n \), and then multiply the angle \( \theta \) by \( n \). This can be expressed mathematically as:

\[(r \text{ cis } \theta)^n = r^n \text{ cis } (n \theta)\]

Here, "cis" is shorthand for \( \cos \theta + i \sin \theta \). For example, if you have a complex number represented as \( 3 \text{ cis } 15^\circ \) and you want to square it, you would calculate:

\[3^2 \text{ cis } (2 \times 15^\circ) = 9 \text{ cis } 30^\circ\]

This method not only simplifies calculations but also provides a quick way to find results without tedious multiplication. For instance, if you need to evaluate \( (4 \text{ cis } \frac{\pi}{6})^3 \), you would compute:

\[4^3 \text{ cis } (3 \times \frac{\pi}{6}) = 64 \text{ cis } \frac{\pi}{2}\]

Using de Moivre's theorem allows for rapid calculations, especially when dealing with higher powers, making it a valuable tool in complex number operations. This theorem is particularly advantageous in scenarios where direct multiplication would be cumbersome, thus streamlining the problem-solving process.

Given $z=3CiS\left(\frac{4\pi}{5}\right)$, calculate $z^5$.

• A. $z^5=243CiS\left(4\pi\right)$
• B. $z^5=243CiS\left(32\pi\right)$
• C. $z^5=243CiS\left(100\right)$
• D. $z^5=243CiS\left(4\right)$

Answer: (A)

Powers Of Complex Numbers In Polar Form (DeMoivre's Theorem) Example 1

To solve the problem of dividing two complex numbers in polar form and raising the result to the third power, we can break the process into two main steps: division and exponentiation.

Given two complex numbers, \( z_1 \) and \( z_2 \), in polar form, we start by finding the quotient \( \frac{z_1}{z_2} \). The division of complex numbers in polar form involves dividing their magnitudes (r values) and subtracting their angles. If we denote the magnitudes of \( z_1 \) and \( z_2 \) as \( r_1 \) and \( r_2 \), and their angles as \( \theta_1 \) and \( \theta_2 \), the formula for division is:

\[\frac{z_1}{z_2} = \frac{r_1}{r_2} \text{cis}(\theta_1 - \theta_2)\]

For example, if \( r_1 = 4 \) and \( r_2 = \frac{1}{2} \), we calculate:

\[\frac{4}{\frac{1}{2}} = 4 \times 2 = 8\]

Next, we subtract the angles. If \( \theta_1 = 25^\circ \) and \( \theta_2 = 10^\circ \), we find:

\[\theta_1 - \theta_2 = 25^\circ - 10^\circ = 15^\circ\]

Thus, the result of the division is:

\[\frac{z_1}{z_2} = 8 \text{cis}(15^\circ)\]

Now, we proceed to the second step, which is raising the result to the third power using De Moivre's theorem. This theorem states that:

\[(z \text{cis}(\theta))^n = r^n \text{cis}(n\theta)\]

Applying this to our result, we have:

\[\left(8 \text{cis}(15^\circ)\right)^3 = 8^3 \text{cis}(3 \times 15^\circ)\]

Calculating \( 8^3 \) gives:

\[8^3 = 512\]

And for the angle:

\[3 \times 15^\circ = 45^\circ\]

Therefore, the final result of \( \left(\frac{z_1}{z_2}\right)^3 \) is:

\[512 \text{cis}(45^\circ)\]

This method illustrates how to handle multiple operations involving complex numbers in polar form by systematically addressing division first and then exponentiation.

Complex Roots

Understanding how to find the roots of complex numbers in polar form is essential for mastering complex analysis. When dealing with roots, such as square roots or cube roots, the process can initially seem tedious, but it becomes manageable with practice. The fundamental concept involves using de Moivre's theorem, which states that for a complex number expressed in polar form as \( r \text{cis} \theta \), raising it to a power or taking a root can be systematically approached.

To find the \( n \)-th root of a complex number, the formula is given by:

\( z_k = r^{\frac{1}{n}} \text{cis} \left( \frac{\theta + 360^\circ k}{n} \right) \)

Here, \( r \) is the modulus of the complex number, \( \theta \) is the argument, and \( k \) takes on integer values from \( 0 \) to \( n-1 \). The term \( 360^\circ k \) accounts for the periodic nature of angles in polar coordinates, allowing for multiple solutions.

For example, to find the cube root of a complex number like \( 8 \text{cis} 45^\circ \), we first calculate the modulus raised to the \( \frac{1}{3} \) power:

\( r^{\frac{1}{3}} = 8^{\frac{1}{3}} = 2 \)

Next, we determine the angles for each root by applying the formula for \( \theta_k \):

\( \theta_k = \frac{45^\circ + 360^\circ k}{3} \)

Calculating for \( k = 0, 1, 2 \) gives:

• For \( k = 0 \): \( \theta_0 = \frac{45^\circ + 360^\circ \cdot 0}{3} = 15^\circ \)
• For \( k = 1 \): \( \theta_1 = \frac{45^\circ + 360^\circ \cdot 1}{3} = 135^\circ \)
• For \( k = 2 \): \( \theta_2 = \frac{45^\circ + 360^\circ \cdot 2}{3} = 255^\circ \)

Thus, the three cube roots of \( 8 \text{cis} 45^\circ \) are:

• \( z_0 = 2 \text{cis} 15^\circ \)
• \( z_1 = 2 \text{cis} 135^\circ \)
• \( z_2 = 2 \text{cis} 255^\circ \)

In summary, finding the roots of complex numbers involves determining the modulus raised to the appropriate power and calculating the angles using the periodic nature of trigonometric functions. By systematically applying these principles, one can efficiently solve for multiple roots of complex numbers.

Complex Roots Example 1

To find the 5th root of the complex number \( z = 1024 \cdot \text{cis}\left(\frac{\pi}{2}\right) \), we start by recognizing that the 5th root can be expressed as \( z^{\frac{1}{5}} \). In polar form, the 5th root of a complex number is calculated using the formula:

\[z^{\frac{1}{n}} = r^{\frac{1}{n}} \cdot \text{cis}\left(\frac{\theta + 2\pi k}{n}\right)\]

Here, \( r \) is the modulus (magnitude) of the complex number, \( \theta \) is the argument (angle), and \( k \) takes on integer values from \( 0 \) to \( n-1 \). In this case, \( r = 1024 \) and \( \theta = \frac{\pi}{2} \), with \( n = 5 \).

First, we calculate the modulus for the 5th root:

\[r^{\frac{1}{5}} = 1024^{\frac{1}{5}} = 4\]

Next, we determine the angles for each of the 5 roots using the formula for \( \theta_k \):

\[\theta_k = \frac{\theta}{n} + \frac{2\pi k}{n}\]

Substituting the known values:

\[\theta_k = \frac{\frac{\pi}{2}}{5} + \frac{2\pi k}{5} = \frac{\pi}{10} + \frac{2\pi k}{5}\]

Now we calculate \( \theta_k \) for \( k = 0, 1, 2, 3, 4 \):

For \( k = 0 \):

\[\theta_0 = \frac{\pi}{10} + 0 = \frac{\pi}{10}\]

For \( k = 1 \):

\[\theta_1 = \frac{\pi}{10} + \frac{2\pi}{5} = \frac{\pi}{10} + \frac{4\pi}{10} = \frac{5\pi}{10} = \frac{\pi}{2}\]

For \( k = 2 \):

\[\theta_2 = \frac{\pi}{10} + \frac{4\pi}{5} = \frac{\pi}{10} + \frac{8\pi}{10} = \frac{9\pi}{10}\]

For \( k = 3 \):

\[\theta_3 = \frac{\pi}{10} + \frac{6\pi}{5} = \frac{\pi}{10} + \frac{12\pi}{10} = \frac{13\pi}{10}\]

For \( k = 4 \):

\[\theta_4 = \frac{\pi}{10} + \frac{8\pi}{5} = \frac{\pi}{10} + \frac{16\pi}{10} = \frac{17\pi}{10}\]

Now we can express each of the 5th roots of \( z \) as:

\[z_k = 4 \cdot \text{cis}(\theta_k)\]

Thus, the 5 solutions are:

\[z_0 = 4 \cdot \text{cis}\left(\frac{\pi}{10}\right)\]

\[z_1 = 4 \cdot \text{cis}\left(\frac{\pi}{2}\right)\]

\[z_2 = 4 \cdot \text{cis}\left(\frac{9\pi}{10}\right)\]

\[z_3 = 4 \cdot \text{cis}\left(\frac{13\pi}{10}\right)\]

\[z_4 = 4 \cdot \text{cis}\left(\frac{17\pi}{10}\right)\]

These represent all the 5th roots of the complex number \( z \). Each root corresponds to a unique angle, demonstrating the periodic nature of complex numbers in polar form.

Here’s what students ask on this topic:

What is DeMoivre's Theorem and how is it used to raise complex numbers to a power?

DeMoivre's Theorem is a formula used to raise complex numbers in polar form to a power. It simplifies the process by allowing you to raise the modulus (r value) to the power and multiply the angle (θ) by the power. The theorem is expressed as: $r^n\mathrm{cis}(n\theta )$. This approach is particularly useful for simplifying calculations involving complex numbers, as it provides a shortcut compared to multiplying the number by itself repeatedly. For example, raising a complex number with r = 3 and θ = 15° to the power of 2 results in 9 cis 30°.

How do you find the cube roots of complex numbers using DeMoivre's Theorem?

Finding cube roots of complex numbers using DeMoivre's Theorem involves determining multiple solutions due to the periodic nature of complex numbers. The general formula is: $r^{\frac{1}{n}}\mathrm{cis}(\frac{\theta }{n}+360k)$, where k ranges from 0 to n-1. For example, the cube root of 8 cis 45° results in three solutions: 15°, 135°, and 255°, corresponding to k values of 0, 1, and 2. This method accounts for the multiple roots that arise from the periodicity of complex numbers.

Why is DeMoivre's Theorem considered a shortcut for raising complex numbers to powers?

DeMoivre's Theorem is considered a shortcut because it simplifies the process of raising complex numbers to powers by using polar form. Instead of multiplying the complex number by itself repeatedly, the theorem allows you to raise the modulus to the power and multiply the angle by the power, expressed as: $r^n\mathrm{cis}(n\theta )$. This approach is faster and more efficient, especially for higher powers, making calculations simpler and quicker.

What are the multiple solutions when finding roots of complex numbers?

When finding roots of complex numbers, multiple solutions arise due to the periodic nature of complex numbers. The general formula for finding roots is: $r^{\frac{1}{n}}\mathrm{cis}(\frac{\theta }{n}+360k)$, where k ranges from 0 to n-1. Each value of k corresponds to a different solution, representing the multiple roots. For example, the cube root of 8 cis 45° results in three solutions: 15°, 135°, and 255°, demonstrating the multiple roots of complex numbers.

How does the periodic nature of complex numbers affect finding their roots?

The periodic nature of complex numbers affects finding their roots by resulting in multiple solutions. This is because complex numbers in polar form can represent multiple angles due to their cyclical nature. When finding roots, the formula $r^{\frac{1}{n}}\mathrm{cis}(\frac{\theta }{n}+360k)$ accounts for this periodicity by including a term for full rotations, allowing for multiple solutions corresponding to different values of k. This results in n distinct roots for an nth root calculation.