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How to calculate the hyperbolic functions of a complex number?

To compute the hyperbolic functions of a complex number, we utilize the definitions of the hyperbolic sine, cosine, and tangent functions.

Let a complex number be defined as:

z=x+iyz = x + iy

where ( x ) and ( y ) are the real and imaginary parts, respectively. The hyperbolic functions of ( z ) are defined as follows:

sinh(z)=ezez2\sinh(z) = \frac{e^z - e^{-z}}{2} cosh(z)=ez+ez2\cosh(z) = \frac{e^z + e^{-z}}{2} tanh(z)=sinh(z)cosh(z)\tanh(z) = \frac{\sinh(z)}{\cosh(z)}

To evaluate these functions, we first express ( e^z ) and ( e^{-z} ) using Euler’s formula:

ez=exeiy=ex(cos(y)+isin(y))e^z = e^x \cdot e^{iy} = e^x \cdot (\cos(y) + i \sin(y)) ez=exeiy=ex(cos(y)isin(y))e^{-z} = e^{-x} \cdot e^{-iy} = e^{-x} \cdot (\cos(y) - i \sin(y))

Now we can substitute these expressions into the definitions of the hyperbolic functions.

Calculating ( \sinh(z) ):

[ \sinh(z) = \frac{e^z - e^{-z}}{2} = \frac{e^x \cdot (\cos(y) + i \sin(y)) - e^{-x} \cdot (\cos(y) - i \sin(y))}{2} ]

This simplifies to:

[ \sinh(z) = \frac{(e^x - e^{-x})}{2} \cos(y) + i \frac{(e^x + e^{-x})}{2} \sin(y) ]

Calculating ( \cosh(z) ):

[ \cosh(z) = \frac{e^z + e^{-z}}{2} = \frac{e^x \cdot (\cos(y) + i \sin(y)) + e^{-x} \cdot (\cos(y) - i \sin(y))}{2} ]

This simplifies to:

[ \cosh(z) = \frac{(e^x + e^{-x})}{2} \cos(y) + i \frac{(e^x - e^{-x})}{2} \sin(y) ]

Calculating ( \tanh(z) ):

Finally, we find ( \tanh(z) ):

[ \tanh(z) = \frac{\sinh(z)}{\cosh(z)} = \frac{\left[ \frac{(e^x - e^{-x})}{2} \cos(y) + i \frac{(e^x + e^{-x})}{2} \sin(y) \right]}{\left[ \frac{(e^x + e^{-x})}{2} \cos(y) + i \frac{(e^x - e^{-x})}{2} \sin(y) \right]} ]

This can be further simplified to:

tanh(z)=sinh(x)cosh(x)+itan(y)\tanh(z) = \frac{\sinh(x)}{\cosh(x)} + i \tan(y)

In summary, the hyperbolic functions for a complex number ( z = x + iy ) can be expressed in terms of its real and imaginary components, allowing for straightforward calculations in complex analysis.

Answered by: Prof. Richard White
A-Level Maths Tutor
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