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Commit d71af17f authored by Timo1104's avatar Timo1104
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Added multiplication image

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1 merge request!2Add lecture on complex numbers properly
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...@@ -159,6 +159,9 @@ $$z^{n} = \left(r e^{{\rm i} \varphi}\right)^{n} = r^{n} e^{{\rm i} n \varphi}$$ ...@@ -159,6 +159,9 @@ $$z^{n} = \left(r e^{{\rm i} \varphi}\right)^{n} = r^{n} e^{{\rm i} n \varphi}$$
$$\sqrt[n]{z} = \sqrt[n]{r e^{{\rm i} \varphi} } = \sqrt[n]{r} e^{{\rm i}\varphi/n} $$ $$\sqrt[n]{z} = \sqrt[n]{r e^{{\rm i} \varphi} } = \sqrt[n]{r} e^{{\rm i}\varphi/n} $$
$$\log(z) = log \left(r e^{{\rm i} \varphi}\right) = log(r) + {\rm i} \varphi$$ $$\log(z) = log \left(r e^{{\rm i} \varphi}\right) = log(r) + {\rm i} \varphi$$
$$z_{1}z_{2} = r_{1} e^{{\rm i} \varphi_{1}} r_{2} e^{{\rm i} \varphi_{2}} = r_{1} r_{2} e^{{\rm i} (\varphi_{1} + \varphi_{2}}$$ $$z_{1}z_{2} = r_{1} e^{{\rm i} \varphi_{1}} r_{2} e^{{\rm i} \varphi_{2}} = r_{1} r_{2} e^{{\rm i} (\varphi_{1} + \varphi_{2}}$$
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)
### Differentiation and integration ### Differentiation and integration
We only consider differentiation and integration over *real* variables. We can then regard the complex ${\rm i}$ as another constant, and use our usual differentiation and integration rules: We only consider differentiation and integration over *real* variables. We can then regard the complex ${\rm i}$ as another constant, and use our usual differentiation and integration rules:
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