diff --git a/src/1_complex_numbers.md b/src/1_complex_numbers.md index fcc89d84ebb7db58b7bea1f2e58931b17e1dce14..6b068e02c825cd418475cabef321c35b508b6bbb 100644 --- a/src/1_complex_numbers.md +++ b/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:  @@ -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