Problem 5 - Complex Analysis for Mathematics and Engineering

Question

Find all the roots in both polar and Cartesian form for each expression. (a) \((-2+2 i)^{8}\). (b) \((-1)^{\frac{1}{3}}\). (c) \((-64)^{\frac{1}{4}}\) (d) \((8)^{\frac{1}{6}}\). (e) \((16 i)^{\frac{1}{2}} .\)

Step-by-Step Solution

Answer

(a) 256; (b) \( \frac{1}{2} \pm \frac{\sqrt{3}}{2}i, -1 \); (c) \( \pm(2 + 2i) \); (d) \( \sqrt{2}, \text{and 5 others} \); (e) \( \pm(2\sqrt{2} + 2\sqrt{2}i) \)."

Step 1: Convert to Polar Form

First, convert (a) \((-2+2i)\) to polar form: Find the modulus \( r = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \) and the argument \( \theta = \tan^{-1}\left(\frac{2}{-2}\right) = \frac{3\pi}{4} \). So, \((-2+2i) = 2\sqrt{2}(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4})\).

Step 2: Calculate the Roots Using De Moivre's Theorem (a)

Consider the expression \((-2+2i)^8\). A root of \( z^n = a \) will have the form \( z = r^{1/n}\left(\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right)\) for \( k = 0, 1, ..., n-1 \). Here, \( n = 8 \), \( r = 2\sqrt{2} \) and \( \theta = \frac{3\pi}{4} \).

Step 3: Evaluate Texponents for (a)

Raise the modulus to calculate the power: \[ (2\sqrt{2})^8 = 256 \]. Multiply the angle by 8 to get the new angle: \( 8 \times \frac{3\pi}{4} = 6\pi \). In polar form: \( 256 (\cos 6\pi + i \sin 6\pi) = 256 (1 + 0i) = 256 \). The Cartesian form is simply 256.

Step 4: Determine Polar Form for the Cube Root (b)

For \((-1)^{\frac{1}{3}}\), in polar form: \((-1) = 1 \times (\cos \pi + i\sin \pi)\). Find the cube roots using \( \theta = \pi / 3 + 2k\pi / 3 \) for \( k = 0, 1, 2\).

Step 5: Calculate Cube Roots (b)

The roots are: for \( k=0: 1(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) = \frac{1}{2} + \frac{\sqrt{3}}{2}i \); for \( k=1: 1(\cos\frac{\pi}{3} + 2\pi / 3 + i\sin\frac{\pi}{3} + 2\pi / 3) = -1 \); for \( k=2: 1(\cos\frac{\pi}{3} + 4\pi / 3 + i\sin\frac{\pi}{3} + 4\pi / 3) = \frac{1}{2} - \frac{\sqrt{3}}{2}i \).

Step 6: Convert -64 and Calculate Fourth Roots (c)

\(-64 = 64 \times (\cos \pi + i\sin \pi)\). The fourth roots are: \( 64^{1/4}(\cos(\frac{\pi}{4}+2k\pi/4) + i\sin(\frac{\pi}{4}+2k\pi/4))\) for \( k = 0, 1, 2, 3 \).

Step 7: Evaluate Fourth Roots (c)

Evaluate \( (64^{1/4} = 2 \sqrt{2}) \). The roots are: \( k=0: 2\sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) = 2 + 2i \); \( k=1: 2\sqrt{2}(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}) = -2 + 2i \); \( k=2: 2\sqrt{2}(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}) = -2 - 2i \); \( k=3: 2\sqrt{2}(\cos\frac{7\pi}{4} + i\sin\frac{7\pi}{4}) = 2 - 2i \).

Step 8: Convert 8 for Sixth Root (d)

12. First: \( 8 = 8 \times (\cos 0 + i\sin 0) \). The sixth roots are: \( 8^{1/6}(\cos(0+2k\pi/6) + i\sin(0+2k\pi/6)) \) for \( k = 0, 1, 2, 3, 4, 5 \).

Step 9: Calculate Sixth Roots (d)

Find \( 8^{1/6} = 2^{1/2} = \sqrt{2} \). The roots are: \( k=0: \sqrt{2}, k=1: \sqrt{2}\times(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) \), etc., calculating all six roots according to the recursive angles.

Step 10: Convert 16i and Calculate Root (e)

\( 16i = 16\times(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2})\). Square roots are: \( 16^{1/2}(\cos(\frac{\pi}{2}/2 + k\pi) + i\sin(\frac{\pi}{2}/2 + k\pi)) \) for \( k = 0, 1 \).

Step 11: Calculate Square Roots (e)

\( 16^{1/2} = 4 \). The roots are: \( k=0: 4(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) = 2 \sqrt{2} + 2 \sqrt{2}i \); and \( k=1: 4(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}) = -2 \sqrt{2} - 2 \sqrt{2}i \).

Key Concepts

Polar Form

Complex numbers can be expressed in both rectangular (Cartesian) form and polar form. While the rectangular form uses coordinates

• Real: represented as 'a'
• Imaginary: represented as 'b'

to denote a complex number as \( a + bi \), polar form expresses them using a distance and angle in the complex plane.
The polar form takes a complex number's modulus, \( r \), and argument, \( \theta \), as parts of the expression. The modulus is the distance of the complex number from the origin, calculated as \( r = \sqrt{a^2 + b^2} \).
The argument, \( \theta \), is the angle the line connecting the origin to the number makes with the positive real axis, found using the inverse tangent: \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).
This allows us to write the complex number in polar form as \( r(\cos \theta + i\sin \theta) \).
By using polar form, operations like multiplication and finding roots become more straightforward, thanks to the magnitude and phase separation.

De Moivre's Theorem

De Moivre's Theorem is an important tool for working with complex numbers in polar form, especially useful in raising complex numbers to powers and extracting roots.
This theorem states that for any complex number with modulus \( r \) and argument \( \theta \), the number raised to the power of \( n \) is expressed as:

• \( [r(\cos \theta + i \sin \theta)]^n = r^n (\cos(n\theta) + i \sin(n\theta)) \)

This makes it easy to compute powers because we simply raise the modulus to the power and multiply the argument by \( n \).
For example, to compute \( (-2 + 2i)^8 \), first convert to polar form, find the modulus \( r = 2\sqrt{2} \) and the argument \( \theta = \frac{3\pi}{4} \). Applying the theorem gives \( (2\sqrt{2})^8(\cos(6\pi) + i\sin(6\pi)) \). This simplifies the operation greatly.
By using De Moivre's Theorem, these tasks simplify to basic algebra involving the modulus and trigonometric angles, without needing intricate calculations in the rectangular form.

Root Calculation

Finding the roots of a complex number is more easily managed using its polar form. When a complex number \( z = r(\cos \theta + i \sin \theta) \) has to be rooted, we find it through:

• \( z^{1/n} = r^{1/n} (\cos(\frac{\theta + 2k\pi}{n}) + i \sin(\frac{\theta + 2k\pi}{n})) \)

where \( k = 0, 1, ..., n-1 \) provide the various roots.
This formula gives us the \( n \) different roots by dividing both the angle and modulus by \( n \). The term \( 2k\pi \) accounts for the complete rotations around the complex plane.
For instance, to find cube roots of \( (-1) \), consider its polar form:

• \( r = 1 \)
• \( \theta = \pi \)

Plugging into the roots formula, the roots come out as combinations of angles modified by \( 2k\pi \), producing solutions like \( \frac{1}{2} + \frac{\sqrt{3}}{2}i \), \(-1\), \( \frac{1}{2} - \frac{\sqrt{3}}{2}i \).
Using the polar form and systematic application of the roots' theorem simplifies what once might have been a more challenging task into straightforward calculation steps, enabling students to grasp and apply complex mathematics efficiently.