Problem 12
Question
$$\cosh (2+3 i)=\cosh (2) \cos (3)+i \sinh (2) \sin (3)=-3.7245+0.5118 i$$
Step-by-Step Solution
Verified Answer
The solution is \( -3.7245 + 0.5118i \). The calculated result matches the given one.
1Step 1: Understanding Complex Cosine Hyperbolic Function
The problem gives the function \( \cosh(z) \) where \( z = 2 + 3i \). \( \cosh(z) \) is defined as \( \cosh(a + bi) = \cosh(a) \cos(b) + i \sinh(a) \sin(b) \). Our task is to compute this expression step by step.
2Step 2: Break Down the Given Function Parts
We first note that \( a = 2 \) and \( b = 3 \). We need to calculate the values of \( \cosh(2) \), \( \cos(3) \), \( \sinh(2) \), and \( \sin(3) \).
3Step 3: Calculate Hyperbolic and Trigonometric Components
Using a calculator:- \( \cosh(2) \approx 3.7622 \)- \( \cos(3) \approx -0.9899 \)- \( \sinh(2) \approx 3.6269 \)- \( \sin(3) \approx 0.1411 \)
4Step 4: Substitute Values into Formula
Insert the calculated values into the equation:\[ \cosh(2+3i) = (3.7622)(-0.9899) + i(3.6269)(0.1411) \]
5Step 5: Calculate the Real Part
Multiply the real components:\( 3.7622 \times -0.9899 \approx -3.7245 \).
6Step 6: Calculate the Imaginary Part
Multiply the imaginary components:\( 3.6269 \times 0.1411 \approx 0.5118 \).
7Step 7: Write Final Answer with Real and Imaginary Parts
Combine the two parts to get the final result: \( \cosh(2+3i) = -3.7245 + 0.5118i \). This confirms the given solution.
Key Concepts
Complex NumbersTrigonometric FunctionsCosine Hyperbolic Function
Complex Numbers
Complex numbers are fascinating mathematical constructs that allow us to solve equations that have no real solutions. They are expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit, defined as \( i = \sqrt{-1} \).
When dealing with functions of complex numbers, such as the hyperbolic cosine function in the exercise above, both the real and imaginary parts contribute to the final result.
- The \( a \) component is known as the real part.
- The \( bi \) component is known as the imaginary part.
When dealing with functions of complex numbers, such as the hyperbolic cosine function in the exercise above, both the real and imaginary parts contribute to the final result.
Trigonometric Functions
Trigonometric functions are essential mathematical tools used to relate the angles of a triangle to the lengths of its sides. These functions include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)), among others.
This allows for calculating the contributions of each component separately, making the manipulations with complex numbers more manageable.
- They are periodic functions, meaning they repeat values over specific intervals.
- Sine and cosine functions have ranges from -1 to 1, making them very useful in modeling wave-like phenomena.
This allows for calculating the contributions of each component separately, making the manipulations with complex numbers more manageable.
Cosine Hyperbolic Function
The cosine hyperbolic function, denoted \( \cosh(x) \), is one of the hyperbolic functions, which are analogs of the ordinary trigonometric functions but for a hyperbola instead of a circle.
It highlights how hyperbolic functions and complex numbers can combine to create intricate and interesting results.
- The formula for \( \cosh(x) \) is \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
- It shares some properties with the regular cosine function, such as being an even function (i.e., \( \cosh(-x) = \cosh(x) \)).
It highlights how hyperbolic functions and complex numbers can combine to create intricate and interesting results.
Other exercises in this chapter
Problem 11
$$e^{\frac{11 \pi}{12} i}=\cos \frac{11 \pi}{12}+i \sin \frac{11 \pi}{12}=-0.9659+0.2588 i$$
View solution Problem 12
$$\cosh ^{-1} i=\ln [(1+\pm \sqrt{2}) i]=\left\\{\begin{array}{ll} \log _{e}(1+\sqrt{2})+\left(\frac{\pi}{2}+2 n \pi\right) i & n=0,\pm 1,\pm 2, \ldots \\ \log
View solution Problem 12
$$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$$
View solution Problem 13
$$e^{-i z}=e^{y-x i}=e^{y} \cos x-i e^{y} \sin x$$
View solution