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Hyperbolic functions

Hyperbolic functions

Hyperbolic functions.

Definition 4.3.1

$\cosh z = \frac 12 (e^{z}+e^{-z})$ .
$\sinh z = \frac {1}{2} (e^{z}-e^{-z})$ .
$\tanh z = \frac {\sinh z}{\cosh z}$ .

Example 4.3.2

$\begin{displaymath}\begin{cases}\cosh i = \frac 12 (e^{i}+e^{-i}) = \frac 12 ( 2... ...= \frac 12 (2i \sin 1 ) = i \sin 1 \approx 0.8415 i \end{cases}\end{displaymath}$

Proposition 4.3.3

$\forall z \in \mathbb{C}, \cosh^2 z - \sinh^2 z =1$ .
$\forall z_1, z_2 \in \mathbb{C}, \sinh (z_1 \pm z_2) = \sinh z_1 \cosh z_2 \pm \cosh z_1 \sinh z_2$
$\forall z_1, z_2 \in \mathbb{C}, \cosh (z_1 \pm z_2) = \cosh z_1 \cosh z_2 \pm \sin z_1h \sinh z_2$
$\begin{displaymath}\begin{cases}\frac{d}{dz} (\sinh z) = \cosh z \ \frac{d}{dz} (\cosh z) = \sinh z \end{cases}\end{displaymath}$
$\forall z \in \mathbb{C}, \begin{cases}\sinh (iz) = i \sin z \ \cosh (iz) = \cos z \end{cases}$ .

Example 4.3.4 (A nice equation) We solve the equation $\sin z =2$ .

Let , where $x,y \in \mathbb{R}$ . Then:

$\displaystyle \sin z = \sin (x+iy) = \sin x \cos (iy) + \sin (iy) \cos x = \sin x \cosh y + i \sinh y \cos x$

We have now:

$\displaystyle \sin z =2 \Longleftrightarrow \begin{cases}\sin x \cosh y =2 \ \sinh y \cos x =0 \end{cases}$
We take care of the second equation:

$\displaystyle \sinh y \cos y =0 \Longleftrightarrow \begin{cases}\sinh y =0 \ ... ...ases}y=0 \ \text{or} \ x=\frac {\pi }{2} + k \pi, k \in \mathbb{Z}\end{cases}$
We substitute into the first equation:
1. $y=0 \Longrightarrow \cosh y = 1 \Longrightarrow \sin x =2$ . This equation has no real solution.
2. $x=\frac {\pi }{2} + 2k \pi \Longrightarrow \cosh y =2$ . This equation is equivalent to
  
  $\displaystyle \frac 12 (e^y+e^{-y})=2$
  
  Its (real) solutions are $\ln (2-\sqrt{3})$ and $\ln (2+\sqrt{3})$ .
3. $x=\frac {\pi }{2} + (2k+1) \pi \Longrightarrow \cosh y =-2$ . This equation has no solution.

In conclusion, the complex solutions of the equation $\sin z =2$ are:

$\displaystyle \frac {\pi }{2} + 2k \pi + i \ln (2-\sqrt{3})$ and $\displaystyle \frac {\pi }{2} + 2k \pi + \ln (2+\sqrt{3})$

Example 4.3.5 (A nice equation - second way) We solve the equation $\sin z =2$ .

$\displaystyle \sin z$	$\displaystyle =2$
$\displaystyle \frac {1}{2i} (e^{iz}-e^{-iz})$	$\displaystyle =2$
$\displaystyle e^{iz}-e^{-iz}$	$\displaystyle =4i$
$\displaystyle e^{2iz} -4ie^{iz}-1$	$\displaystyle =0.$

We substitute $t=e^{iz}$ and solve for

the equation

. The solutions are: $t_1=(2-\sqrt{3})i$ and $t_1=(2+\sqrt{3})i$ . In order to find the corresponding values of

, we need logarithms. They will be defined in the next paragraph (v.i. 4). This exercise will be finished in example 4.6.