Question

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

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

Short answer

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

Step-by-step solution

Given information

Thegiven the function sin h z= sin h x cos y +i cosh x sin 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$\mathit{r}{\mathit{e}}^{\mathbf{i}\mathbf{\theta }}\mathbf{}\mathbf{=}\mathit{r}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{+}\mathit{i}\mathit{r}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }$.

 Use exponential and polar form to expand the equation

The exponential form of the given equation is,

$\sin hz=\frac{e^z-e^{-z}}{2}$ …. (1)

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

$\sin hz=\frac{e^{\left(x+yi\right)}-e^{\left(-x+yi\right)}}{2}$

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

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

Convert exponential form into polar form.

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

Replace cos xi by cosh xand sin xi by i sinh x

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

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