Problem 22
Question
$$(2 \sqrt{2})^{5}\left[\cos \left(-\frac{5 \pi}{4}\right)+i \sin \left(-\frac{5 \pi}{4}\right)\right]=-128+128 i$$
Step-by-Step Solution
Verified Answer
The result of the expression is \(-128 + 128i\).
1Step 1: Convert to Polar Form
The given expression is in the form \( w^5 \), where \( w = 2\sqrt{2}(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4})) \). This is a polar representation with \( r = 2\sqrt{2} \) and \( \theta = -\frac{\pi}{4} \).
2Step 2: Apply De Moivre's Theorem
Using De Moivre's Theorem, \[ (r(\cos \theta + i \sin \theta))^n = r^n (\cos(n \theta) + i \sin(n \theta)) \]. Here, \( r = 2\sqrt{2} \), \( \theta = -\frac{\pi}{4} \), and \( n = 5 \).
3Step 3: Calculate \( r^n \)
Calculate \( r^5 \): \( (2\sqrt{2})^5 = 2^5 \times (\sqrt{2})^5 = 32 \times 4\sqrt{2} = 128 \).
4Step 4: Calculate \( n\theta \)
Find \( 5 \times (-\frac{\pi}{4}) = -\frac{5\pi}{4} \) and then adjust it to the equivalent angle within the range \( [0, 2\pi) \). The equivalent angle is \( \frac{3\pi}{4} \) as \( -\frac{5\pi}{4} + 2\pi = \frac{3\pi}{4} \).
5Step 5: Formulate the Result
Substitute \( r^5 \) and \( n\theta \) into De Moivre's theorem: \[ 128 \left( \cos \left( \frac{3\pi}{4} \right) + i \sin \left( \frac{3\pi}{4} \right) \right) \].
6Step 6: Evaluate Trigonometric Functions
Evaluate \( \cos \left( \frac{3\pi}{4} \right) = -\frac{\sqrt{2}}{2} \) and \( \sin \left( \frac{3\pi}{4} \right) = \frac{\sqrt{2}}{2} \).
7Step 7: Compute Final Result
Calculate \[ 128 \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = -128 + 128i \].
8Step 8: Verify the Answer
The calculated result \( -128 + 128i \) matches the given expression, confirming the solution is correct.
Key Concepts
Polar FormTrigonometric FunctionsComplex Numbers
Polar Form
Polar form is an elegant way to express complex numbers, which are numbers consisting of a real part and an imaginary part. In polar form, these numbers are represented in terms of a magnitude (often called 'radius') and an angle. This is particularly useful in multiplying or raising complex numbers to powers.
When a complex number is in polar form, it looks something like this:
When a complex number is in polar form, it looks something like this:
- The magnitude or modulus, usually denoted as \( r \), represents the distance from the origin to the point in the complex plane.
- The angle, sometimes called the argument, denoted as \( \theta \), is the angle formed with the positive x-axis.
- Thus, the polar form of a complex number is written as \( r (\cos \theta + i \sin \theta) \).
Trigonometric Functions
Trigonometric functions, namely cosine and sine, play a significant role in expressing the polar form of complex numbers. These functions help define the exact position of the complex number on the complex plane based on its angle.
- The cosine function, \( \cos \theta \), determines the horizontal position (real part) of the complex number.
- The sine function, \( \sin \theta \), determines the vertical position (imaginary part) of the complex number.
- In essential trigonometry, - \( \cos(\theta) = \frac{adjacent}{hypotenuse} \), and - \( \sin(\theta) = \frac{opposite}{hypotenuse} \).
Complex Numbers
Complex numbers are an extension of real numbers that include a special unit called 'i' which represents the square root of -1. A complex number, therefore, has both a real and an imaginary part, expressed as \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit.
- Real part (\( a \)): This part is similar to ordinary real numbers you are familiar with.
- Imaginary part (\( bi \)): This part involves the imaginary unit where \( i^2 = -1 \).
- The combination of these parts allows complex numbers to perform functions such as solving equations where ordinary real numbers fall short.
Other exercises in this chapter
Problem 22
$$\lim _{z \rightarrow 1+i} \frac{z^{2}-2 z+2}{z^{2}-2 i}=\lim _{z \rightarrow 1+i} \frac{[z-(1+i)][z-(1-i)]}{[z-(1+i)][z-(-1-i)]}=\frac{1}{2}+\frac{1}{2} i$$
View solution Problem 22
(a) \(u=e^{x^{2}-y^{2}} \cos 2 x y, \quad v=e^{x^{2}-y^{2}} \sin 2 x y ; \quad \frac{\partial u}{\partial x}=-2 y e^{x^{2}-y^{2}} \sin 2 x y+2 x e^{x^{2}-y^{2}}
View solution Problem 22
$$[(1+i)(1-i)]^{2}(1-i)=4-4 i$$
View solution Problem 23
Along the \(y\) -axis,\(\lim _{z \rightarrow 0} \frac{x+i y}{x-i y}=\lim _{y \rightarrow 0} \frac{i y}{-i y}=-1,\) whereas along the \(x\) -axis,\(\lim _{z \rig
View solution