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Math  /  Algebra

Question3. Prove the following identities: (a) sinh⁡(z+π)=−sinh⁡(z)\sinh (z+\pi)=-\sinh (z) (b) cosh⁡(z)=cosh⁡(x)cos⁡(y)+isinh⁡(x)sin⁡(y)\cosh (z)=\cosh (x) \cos (y)+i \sinh (x) \sin (y)

Studdy Solution

STEP 1

1. We are working with hyperbolic functions and complex numbers.
2. The hyperbolic sine and cosine functions are defined as: - sinh⁡(z)=ez−e−z2\sinh(z) = \frac{e^z - e^{-z}}{2} - cosh⁡(z)=ez+e−z2\cosh(z) = \frac{e^z + e^{-z}}{2}
3. The complex number zz can be expressed as z=x+iyz = x + iy, where xx and yy are real numbers.

STEP 2

1. Prove identity (a): sinh⁡(z+π)=−sinh⁡(z)\sinh(z+\pi) = -\sinh(z).
2. Prove identity (b): cosh⁡(z)=cosh⁡(x)cos⁡(y)+isinh⁡(x)sin⁡(y)\cosh(z) = \cosh(x) \cos(y) + i \sinh(x) \sin(y).

STEP 3

Start with the definition of sinh⁡(z)\sinh(z) and apply it to sinh⁡(z+π)\sinh(z+\pi):
sinh⁡(z+π)=ez+π−e−(z+π)2\sinh(z+\pi) = \frac{e^{z+\pi} - e^{-(z+\pi)}}{2}
Use the property of exponents ez+π=ez⋅eπe^{z+\pi} = e^z \cdot e^{\pi} and e−(z+π)=e−z⋅e−πe^{-(z+\pi)} = e^{-z} \cdot e^{-\pi}.

STEP 4

Substitute the exponential expressions:
sinh⁡(z+π)=ez⋅eπ−e−z⋅e−π2\sinh(z+\pi) = \frac{e^z \cdot e^{\pi} - e^{-z} \cdot e^{-\pi}}{2}
Since eπ=−1e^{\pi} = -1 and e−π=−1e^{-\pi} = -1, simplify:
sinh⁡(z+π)=ez⋅(−1)−e−z⋅(−1)2=−ez+e−z2=−ez−e−z2=−sinh⁡(z)\sinh(z+\pi) = \frac{e^z \cdot (-1) - e^{-z} \cdot (-1)}{2} = \frac{-e^z + e^{-z}}{2} = -\frac{e^z - e^{-z}}{2} = -\sinh(z)

STEP 5

Express z=x+iyz = x + iy and substitute into cosh⁡(z)\cosh(z):
cosh⁡(z)=cosh⁡(x+iy)=ex+iy+e−(x+iy)2\cosh(z) = \cosh(x + iy) = \frac{e^{x+iy} + e^{-(x+iy)}}{2}
Use the properties of exponents to expand:
ex+iy=ex⋅eiyande−(x+iy)=e−x⋅e−iye^{x+iy} = e^x \cdot e^{iy} \quad \text{and} \quad e^{-(x+iy)} = e^{-x} \cdot e^{-iy}

STEP 6

Apply Euler's formula: eiy=cos⁡(y)+isin⁡(y)e^{iy} = \cos(y) + i\sin(y) and e−iy=cos⁡(y)−isin⁡(y)e^{-iy} = \cos(y) - i\sin(y):
cosh⁡(z)=ex(cos⁡(y)+isin⁡(y))+e−x(cos⁡(y)−isin⁡(y))2\cosh(z) = \frac{e^x (\cos(y) + i\sin(y)) + e^{-x} (\cos(y) - i\sin(y))}{2}
Combine terms:
=(ex+e−x)cos⁡(y)+i(ex−e−x)sin⁡(y)2= \frac{(e^x + e^{-x})\cos(y) + i(e^x - e^{-x})\sin(y)}{2}

STEP 7

Recognize the hyperbolic identities:
ex+e−x=2cosh⁡(x)andex−e−x=2sinh⁡(x)e^x + e^{-x} = 2\cosh(x) \quad \text{and} \quad e^x - e^{-x} = 2\sinh(x)
Substitute these into the expression:
cosh⁡(z)=cosh⁡(x)cos⁡(y)+isinh⁡(x)sin⁡(y)\cosh(z) = \cosh(x)\cos(y) + i\sinh(x)\sin(y)
The identities have been proven:
(a) sinh⁡(z+π)=−sinh⁡(z)\sinh(z+\pi) = -\sinh(z)
(b) cosh⁡(z)=cosh⁡(x)cos⁡(y)+isinh⁡(x)sin⁡(y)\cosh(z) = \cosh(x) \cos(y) + i \sinh(x) \sin(y)

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