Q12P - Mathematical Methods in Physical Sciences

Question

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

 $\mathbf{cos}\mathbf{ }\mathbf{z}\mathbf{=}\mathbf{cos}\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{cosh}\mathbf{ }\mathbf{y}\mathbf{-}\mathbf{i}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{sinh}\mathbf{ }\mathbf{y}$

Step-by-Step Solution

Answer

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

Step 1: Given information

 The given function cos z=cos x cosh y- i sin x sinh 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,

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

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

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

 Convert exponential form into polar form.

$\cos z=\frac{\left[\cos x+i \sin x\right] \left[\cos yi+i \sin yi\right]+\left[\cos x-i \sin x\right] \left[\cos yi-i \sin yi\right]}{2}$                             …(2)

Step 4: Replace cos yi by cosh y and sin yi by sinh y

Replace cos yi by cosh y and sin yi by sinh y in equation (2).

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