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  

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

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. $$\begin{cases}\frac{d}{dz} (\sinh z) = \cosh z \ \frac{d}{dz} (\cosh z) = \sinh z \end{cases}$$
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. $$\sin z = \sin (x+iy) = \sin x \cos (iy) + \sin (iy) \cos x = \sin x \cosh y + i \sinh y \cos x$$

   
   
   We have now:

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

   
   

2. We take care of the second equation:

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

   
   

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

      $$\frac 12 (e^y+e^{-y})=2$$

      
      
      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:

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

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

$$\sin z =2$$

$$\frac {1}{2i} (e^{iz}-e^{-iz}) =2$$

$$e^{iz}-e^{-iz} =4i$$

$$e^{2iz} -4ie^{iz}-1 =0.$$

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.