Hyperbolic functions.

Definition 4.3.1       

  1. $ \cosh z = \frac 12 (e^{z}+e^{-z})$ .
  2. $ \sinh z = \frac {1}{2} (e^{z}-e^{-z})$ .
  3. $ \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       

  1. $ \forall z \in \mathbb{C}, \cosh^2 z - \sinh^2 z =1$ .
  2. $ \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$
  3. $ \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$
  4. \begin{displaymath}\begin{cases}\frac{d}{dz} (\sinh z) = \cosh z \ \frac{d}{dz}
(\cosh z) = \sinh z \end{cases}\end{displaymath}
  5. $ \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 $ z=x+iy$ , where $ x,y \in \mathbb{R}$ . Then:

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

    We have now:

    $\displaystyle \sin z =2 \Longleftrightarrow \begin{cases}\sin x \cosh y =2 \ \sinh y \cos x =0 \end{cases}$    

  2. We take care of the second equation:

    $\displaystyle \sinh y \cos y =0 \Longleftrightarrow \begin{cases}\sinh y =0 \ ...
...ases}y=0 \ \text{or} \ x=\frac {\pi }{2} + k \pi, k \in \mathbb{Z}\end{cases}$    

  3. 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 $ t$ the equation $ t^2-4it-1=0$ . The solutions are: $ t_1=(2-\sqrt{3})i$ and $ t_1=(2+\sqrt{3})i$ . In order to find the corresponding values of $ z$ , we need logarithms. They will be defined in the next paragraph (v.i. 4). This exercise will be finished in example  4.6.

Noah Dana-Picard 2007-12-24