Added example of z^4=1,

removed hyperbolic functions

src/1_complex_numbers.md

@@ -158,7 +158,7 @@ Some operations which are common in real analysis are then easily derived for th
 $$z^{n} = \left(r e^{{\rm i} \varphi}\right)^{n} = r^{n} e^{{\rm i} n \varphi}$$
 $$\sqrt[n]{z} = \sqrt[n]{r e^{{\rm i} \varphi} } = \sqrt[n]{r} e^{{\rm i}\varphi/n} $$
 $$\log(z) = log \left(r e^{{\rm i} \varphi}\right) = log(r) + {\rm i} \varphi$$
-$$z_{1}z_{2} = r_{1} e^{{\rm i} \varphi_{1}} r_{2} e^{{\rm i} \varphi_{2}} = r_{1} r_{2} e^{{\rm i} (\varphi_{1} + \varphi_{2}}$$
+$$z_{1}z_{2} = r_{1} e^{{\rm i} \varphi_{1}} r_{2} e^{{\rm i} \varphi_{2}} = r_{1} r_{2} e^{{\rm i} (\varphi_{1} + \varphi_{2}})$$
 We see that during multiplication, the norm of the new number is the *product* of the norms of the multiplied numbers, and its argument is the *sum* of the arguments of the multiplied numbers. In the complex plane, this looks as follows:
 
 ![image](figures/complex_numbers_12_0.svg)
@@ -168,6 +168,14 @@ We only consider differentiation and integration over *real* variables. We can t
 $$\frac{d}{d\varphi} e^{{\rm i} \varphi} = e^{{\rm i} \varphi} \frac{d}{d\varphi} ({\rm i} \varphi) ={\rm i} e^{{\rm i} \varphi} .$$
 $$\int_{0}^{\pi} e^{{\rm i} \varphi} = \frac{1}{{\rm i}} \left[ e^{{\rm i} \varphi} \right]_{0}^{\pi} = -{\rm i}(-1 -1) = 2 {\rm i}$$
 
+**Example** Find all solutions solving $z^4 = 1$. 
+Of course, we know that $z = \pm 1$ are two solutions, but which other solutions are possible? We take a systematic approach:
+$$ z = e^{{\rm i} \varphi} \Rightarrow z^4 = e^{4{\rm i} \varphi} = 1 $$
+$$\Leftrightarrow 4 \varphi = n 2 \pi$$
+$$\Leftrightarrow \varphi = 0, \varphi = \frac{\pi}{2}, \varphi = -\frac{\pi}{2}, \varphi = \pi$$
+$$\Leftrightarrow z = 1, z = i, z = -i, z = -1$$
+
+
 
 Let us show some tricks where the simple properties of the exponential
 function helps in re-deriving trigonometric identities.
@@ -177,11 +185,11 @@ function helps in re-deriving trigonometric identities.
     $z_i = \exp({\rm i} \varphi_i)$, $i=1, 2$. Then:
     $$z_1 z_2 = \exp[{\rm i} (\varphi_1 + \varphi_2)].$$ The left hand
     side can be written as
-    /begin{align}
+    $$/begin{align}
     z_1 z_2 & = \left[ \cos(\varphi_1) + {\rm i} \sin(\varphi_1) \right] \left[ \cos(\varphi_2) + {\rm i} \sin(\varphi_2) \right] \\
     & = \cos\varphi_1 \cos\varphi_2 - \sin\varphi_1 \sin\varphi_2 + {\rm i} \left( \cos\varphi_1 \sin\varphi_2 + 
     \sin\varphi_1 \cos\varphi_2 \right).
-    \end{align}
+    \end{align}$$
     On the other hand, the right
     hand side can be written as
     $$\exp[{\rm i} (\varphi_1 + \varphi_2)] = \cos(\varphi_1 + \varphi_2) + {\rm i} \sin(\varphi_1 + \varphi_2).$$
@@ -204,38 +212,6 @@ function helps in re-deriving trigonometric identities.
     and imaginary parts leads to $$\cos'\varphi = - \sin\varphi;$$
     $$\sin'\varphi = \cos\varphi.$$
 
-### Hyperbolic functions
-
-
-From
-$e^{\rm i \varphi} = \left( \cos\varphi + {\rm i} \sin\varphi\right)$,
-it immediately follows that
-$$\cos\varphi = \frac{e^{{\rm i} \varphi} + e^{-{\rm i} \varphi}}{2}.$$
-and
-$$\sin\varphi = \frac{e^{{\rm i} \varphi} - e^{-{\rm i} \varphi}}{2{\rm i}}.$$
-It is then tempting to generalise these functions for imaginary angles.
-These functions are known as hyperbolic functions. They are are called
-the hyperbolic cosine and hyperbolic sine functions and they are denoted
-as $\sinh$ and $\cosh$: $$\cosh(x) = \frac{e^x + e^{-x}}{2};$$
-$$\sinh(x) = \frac{e^x - e^{-x}}{2}.$$ From these definitions the
-following properties can easily be derived.
-
-1.  Derivatives $$\frac{d\cosh(x)}{dx} = \sinh(x);$$
-    $$\frac{d\sinh(x)}{dx} = \cosh(x).$$
-
-2.  $$\cosh^2(x) - \sinh^2(x) = 1.$$
-
-3.  ‘Double angle’ formulas: $$\cosh(2x) = \cosh^2(x) + \sinh^2(x);$$
-    $$\sinh(2x) = 2\cosh(x) \sinh(x).$$
-
-It may seem that these function are rather exotic; however they occur in
-everyday life: the shapes of power lines and of soap films can be
-described by hyperbolic cosines and sines!
-
-Finally, the hyperbolic tangent is defined as
-$$\tanh(x) = \frac{\sinh(x)}{\cosh(x)}.$$ Its derivative is given as
-$$\tanh'(x) = 1 + \frac{\sinh^2 x}{\cosh^2 (x)} = - \frac{1}{\cosh^2(x)}.$$
-
 ## Summary
 
 -   A complex number $z$ has the form $$z = a + b \rm i$$ where $a$ and