Q4P - Mathematical Methods in Physical Sciences

Question

Verify each of the following by using equations (11.4), (12.2), and (12.3).

cosh z=cosh x cos y+i sinh x sin y

Step-by-Step Solution

Answer

The equation cosh z=cosh x cos y+i sinh x sin y is verified using the equations (11.4), (12.2) and (12.3).

Step 1: Given information

The given function is cosh z=cosh x cos y+i sin h x sin y.

Step 2: Definition of Hyperbolic Function.

A hyperbolic function is a representation of the relationship between a point's distances from the origin to the coordinate axes as a function of an angle.

Relation between the exponential and polar form is $\mathit{r}{\mathit{e}}^{\mathbf{i}\mathbf{\theta }}\mathbf{=}\mathit{r}\mathbf{ }\mathit{c}\mathit{o}\mathit{s}\mathbf{ }\mathit{\theta }\mathbf{+}\mathit{i}\mathit{r}\mathbf{ }\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\mathit{\theta }$.

Step 3: Use exponential and polar form to expand the equation

The exponential form of the given equation is,

 $\cos h z=\frac{e^z+e^{-z}}{2}$                                                                                                         ...(1)

 

Let $z=x+yi$and put in equation (1).

$\cosh \mathrm{z}=\frac{{\mathrm{e}}^{\left(\mathrm{x}+\mathrm{yi}\right)}+{\mathrm{e}}^{\left(-\mathrm{x}-\mathrm{yi}\right)}}{2}$

 

Write x+yi as i(xi+y).

$$\begin{gathered} \cos h z=\frac{e^{\left[\left(-xi+y\right)i\right]}+e^{\left[\left(xi-y\right)i\right]}}{2} \\ \cos h z=\frac{e^{\left(-xi\right)i}.e^{\left(yi\right)}+e^{\left(xi\right)i}.e^{\left(-yi\right)}}{2} \end{gathered}$$

 

Convert exponential form into polar form.

$\cos h z=\frac{\left[\cos xi-i \sin ix\right] \left[\cos y+i \sin y\right]+\left[\cos xi+i \sin ix\right] \left[\cos y-i \sin y\right]}{2}$                  …. (2)

Step 4:Replace cos xi by cosh x and sin xi by i sinh x

Replace cos xi by cosh x and sin xi by i sinh x  in equation (2).

$$\begin{gathered} \cos h z=\frac{\left[\cos h x+\sin h x \right] \left[\cos y+i \sin y\right]+\left[\cos h x+\sin h x \right] \left[\cos y+i \sin y\right]}{2} \\ =\left\{\begin{array}{c}\frac{\left[\cosh \mathrm{x}.\cos \mathrm{y} +\mathrm{i} \sinh \mathrm{x}. \sin \mathrm{y}+\mathrm{i} \cosh \mathrm{x}. \sin \mathrm{y}+\sinh \mathrm{x}. \cos \mathrm{y}\right]}{2}\\ +\frac{\left[\cosh \mathrm{x}.\cos \mathrm{y} +\mathrm{i} \sinh \mathrm{x}. \sin \mathrm{y}-\mathrm{i} \cosh \mathrm{x}. \sin \mathrm{y}-\sinh \mathrm{x}. \cos \mathrm{y}\right]}{2}\end{array}\right\} \\ \mathrm{Cancel} \mathrm{similar} \mathrm{terms}. \\ \cosh \mathrm{z}=\frac{2 \cosh \mathrm{x} \cos \mathrm{y}}{2} \frac{2\mathrm{i} \sin \mathrm{y} \cosh \mathrm{x} }{2} \\ \cosh \mathrm{z}=\sin \mathrm{h} \mathrm{x} \cos \mathrm{y}+\mathrm{i} \sin \mathrm{y} \cosh \mathrm{x} \\ \mathrm{Hence} \mathrm{the} \mathrm{equation} \mathrm{is} \mathrm{verified}. \end{gathered}$$