Problem 12
Question
$$e^{5+\frac{5 \pi}{2} i}=e^{5} \cos \frac{5 \pi}{2}+i e^{5} \sin \frac{5 \pi}{2}=e^{5} i$$
Step-by-Step Solution
Verified Answer
The expression simplifies to \( e^5 i \).
1Step 1: Understand the Problem
The given expression involves Euler's formula for complex exponentials. We need to compute \( e^{5 + \frac{5\pi}{2} i} \) and verify the equality.
2Step 2: Apply Euler's Formula
Euler's formula states that \( e^{ix} = \cos(x) + i\sin(x) \). For an expression of the form \( e^{a + bi} \), it equals \( e^a(\cos(b) + i\sin(b)) \). Here, \( a = 5 \) and \( b = \frac{5\pi}{2} \).
3Step 3: Calculate the Real Part
Using \( \cos(x) = \cos(x + 2\pi k) \), we simplify \( \cos\left(\frac{5\pi}{2}\right) \). It's equivalent to \( \cos\left(\frac{\pi}{2}\right) \). Hence, \( \cos\left(\frac{5\pi}{2}\right) = 0 \).
4Step 4: Calculate the Imaginary Part
Similarly, \( \sin(x) = \sin(x + 2\pi k) \), thus \( \sin\left(\frac{5\pi}{2}\right) \) simplifies to \( \sin\left(\frac{\pi}{2}\right) \). Thus, \( \sin\left(\frac{5\pi}{2}\right) = 1 \).
5Step 5: Write the Complex Number
Plug the values back into the Euler's formula expression: \( e^{5} \left( \cos\left(\frac{5\pi}{2}\right) + i\sin\left(\frac{5\pi}{2}\right) \right) = e^{5}(0 + i\cdot 1) = e^5 i \).
Key Concepts
Euler's FormulaComplex NumbersTrigonometric Functions
Euler's Formula
Euler's Formula is an important and fascinating equation in mathematics. It connects the world of complex numbers and trigonometric functions. The formula is expressed as \( e^{ix} = \cos(x) + i\sin(x) \). This is particularly useful in the analysis of complex exponential functions.
- The formula shows how complex exponentials relate to trigonometric functions.
- It forms the basis for tackling problems involving complex numbers and their exponential forms.
Complex Numbers
Complex numbers extend the idea of one-dimensional numbers (real numbers) to two dimensions. This introduction involves the imaginary unit \( i \), where \( i^2 = -1 \). Complex numbers are typically represented as \( a + bi \).
- Here, \( a \) is the real part.
- \( b \) is the imaginary part.
- Any complex number can be expressed in polar form as \( r(\cos(\theta) + i\sin(\theta)) \), which Euler's formula transforms into \( re^{i\theta} \).
- This representation simplifies multiplication and division of complex numbers.
Trigonometric Functions
Trigonometric functions, including sine and cosine, are fundamental in analyzing any periodic or oscillatory processes. Their crucial property is periodicity, meaning these functions repeat their values in regular intervals, called periods.
- The function \( \cos(x) \) repeats every \( 2\pi \) radians, i.e., \( \cos(x + 2\pi k) = \cos(x) \) for any integer \( k \).
- Similarly, \( \sin(x) \) is also periodic with period \( 2\pi \), i.e., \( \sin(x + 2\pi k) = \sin(x) \).
Other exercises in this chapter
Problem 12
$$\cosh ^{-1} i=\ln [(1+\pm \sqrt{2}) i]=\left\\{\begin{array}{ll} \log _{e}(1+\sqrt{2})+\left(\frac{\pi}{2}+2 n \pi\right) i & n=0,\pm 1,\pm 2, \ldots \\ \log
View solution Problem 12
$$\cosh (2+3 i)=\cosh (2) \cos (3)+i \sinh (2) \sin (3)=-3.7245+0.5118 i$$
View solution Problem 13
$$e^{-i z}=e^{y-x i}=e^{y} \cos x-i e^{y} \sin x$$
View solution Problem 14
$$\cos \left(\frac{\pi}{2}+i \ln 2\right)=\cos \frac{\pi}{2} \cosh (\ln 2)-i \sin \frac{\pi}{2} \sinh (\ln 2)=-i \cdot \frac{e^{\ln 2}-e^{\ln 2^{-1}}}{2}=-i \cd
View solution