Problem 29
Question
$$ \text { Prove that } \cosh ^{2} z-\sinh ^{2} z=1 \text {. } $$
Step-by-Step Solution
Verified Answer
The identity \(\cosh^2 z - \sinh^2 z = 1\) holds true, as proven with hyperbolic function definitions.
1Step 1: Recall Definitions
Recall the definitions of hyperbolic functions: \[\cosh z = \frac{e^z + e^{-z}}{2} \] and \[\sinh z = \frac{e^z - e^{-z}}{2}.\] These definitions will be important for proving the identity.
2Step 2: Apply Cosine Hyperbolic to the Left Side
Calculate \(\cosh^2 z\) by squaring the definition of hyperbolic cosine: \[\cosh^2 z = \left(\frac{e^z + e^{-z}}{2}\right)^2.\] Expanding the square: \[\cosh^2 z = \frac{e^{2z} + 2 + e^{-2z}}{4}.\]
3Step 3: Apply Sine Hyperbolic to the Left Side
Calculate \(\sinh^2 z\) by squaring the definition of hyperbolic sine: \[\sinh^2 z = \left(\frac{e^z - e^{-z}}{2}\right)^2.\] Expanding the square: \[\sinh^2 z = \frac{e^{2z} - 2 + e^{-2z}}{4}.\]
4Step 4: Substitute and Simplify
Substitute \(\cosh^2 z\) and \(\sinh^2 z\) into the given identity \(\cosh^2 z - \sinh^2 z = 1\):\[\frac{e^{2z} + 2 + e^{-2z}}{4} - \frac{e^{2z} - 2 + e^{-2z}}{4}.\]Simplify the expression by combining terms: \[= \frac{(e^{2z} + 2 + e^{-2z}) - (e^{2z} - 2 + e^{-2z})}{4} = \frac{4}{4} = 1.\]
5Step 5: Conclusion
Thus, the identity \(\cosh^2 z - \sinh^2 z = 1\) is proven by substituting definitions and simplifying the expressions.
Key Concepts
Hyperbolic IdentitiesHyperbolic CosineHyperbolic Sine
Hyperbolic Identities
Hyperbolic identities are analogous to trigonometric identities, but they relate to hyperbolic functions instead of trigonometric ones. One of the basic hyperbolic identities is \( \cosh^2 z - \sinh^2 z = 1 \). This identity is fundamental because it resembles the Pythagorean identity in trigonometry, where \( \cos^2 \theta + \sin^2 \theta = 1 \).
However, unlike trigonometric functions, hyperbolic functions are defined using exponential functions and are relevant in hyperbolic geometry and complex analysis.
Here’s how to prove this identity:
However, unlike trigonometric functions, hyperbolic functions are defined using exponential functions and are relevant in hyperbolic geometry and complex analysis.
Here’s how to prove this identity:
- Recall the definitions \( \cosh z = \frac{e^z + e^{-z}}{2} \) and \( \sinh z = \frac{e^z - e^{-z}}{2} \).
- Square both expressions to find \( \cosh^2 z \) and \( \sinh^2 z \).
- Subtract the squared values to establish the hyperbolic identity.
Hyperbolic Cosine
The hyperbolic cosine function, denoted as \( \cosh z \), is defined by the formula \( \cosh z = \frac{e^z + e^{-z}}{2} \). This function describes the shape of an idealized hanging cable or chain, which is why it is often associated with a catenary curve.
Its properties are similar to those of the usual cosine function but are distinguished by its definition in terms of exponential functions.
Some key features of the hyperbolic cosine include:
Its properties are similar to those of the usual cosine function but are distinguished by its definition in terms of exponential functions.
Some key features of the hyperbolic cosine include:
- Even Function: \( \cosh(-z) = \cosh(z) \), meaning it is symmetric about the \( y \)-axis.
- Always Positive: \( \cosh(z) > 0 \) for all real numbers \( z \).
- Approaches infinity as \( z \) approaches infinity in both directions.
Hyperbolic Sine
The hyperbolic sine function, denoted as \( \sinh z \), is defined as \( \sinh z = \frac{e^z - e^{-z}}{2} \). This function, much like its trigonometric counterpart, illustrates a relationship with hyperbolas that is equivalent to that of the sine function with circles.
The graphs of \( \sinh z \) portray an increasing nature, similar to the sine but unbounded and more linear.
Key properties of hyperbolic sine are:
The graphs of \( \sinh z \) portray an increasing nature, similar to the sine but unbounded and more linear.
Key properties of hyperbolic sine are:
- Odd Function: \( \sinh(-z) = -\sinh(z) \), indicating symmetry about the origin.
- Range: Values can be both positive and negative, and \( \sinh(z) \) ranges from \(-\infty\) to \(\infty\).
- The slope at the origin is 1, and it increases exponentially for large \( z \).
Other exercises in this chapter
Problem 29
In Problems 29-34, express \(\operatorname{Ln} z\) in the form \(a+i b\). \(z=6-6 i\)
View solution Problem 29
$$ \text { If } f(z)=z^{24}-3 z^{20}+4 z^{12}-5 z^{6}, \text { find } f\left(\frac{1+i}{\sqrt{2}}\right) $$
View solution Problem 29
In Problems 27-32, let \(z=x+i y\). Find the indicated expression. $$ \operatorname{Im}(2 z+4 \bar{z}-4 i) $$
View solution Problem 30
Use (10) to compute all roots. Sketch these roots on an appropriate circle centered at the origin. $$ (-1+i)^{1 / 3} $$
View solution