Question

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

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

Short 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-by-step solution

Given information

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

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 $\mathit{r}{\mathit{e}}^{\mathbf{i}\mathbf{\theta }}\mathbf{=}\mathit{r}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{+}\mathit{i}\mathit{r}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }$.

 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)

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).

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