55 lines
1.7 KiB
TeX
55 lines
1.7 KiB
TeX
\documentclass[a5paper,10pt]{article}
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\usepackage{pl}
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\usepackage[margin=1cm]{geometry}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage[utf8]{inputenc}
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\usepackage{polski}
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\title{Trygonometria i liczby zespolone \\
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\large Algorytmy kwantowe}
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\date{2021-02-27}
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\author{Robert Bendun}
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\newcommand{\mi}{\mathrm{i}}
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\begin{document}
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\maketitle
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\section{Trygonometria}
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\section{Liczby zespolone}
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\subsection{Postać algebraiczna}
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\begin{description}
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\item $ \alpha \pm \beta = \left( a + b\mi \right) \pm \left( c + d\mi \right)
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= \left( a \pm c \right) + \left( b \pm d \right)\mi$
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\item $ \alpha\beta = \left( a + b\mi \right) \left( c + d\mi \right)
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= \left( ac - bd \right) + \left( bc + ad \right)\mi$
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\item $ \frac{\alpha}{\beta} = \frac{a + b\mi}{c + d\mi} = \frac{(ac + bd) + (bc - ad)\mi}{c^2 + d^2} $
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\item[Norma] $ |\alpha| = |a + b\mi| = \sqrt{a^2 + b^2} $
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\item[Sprzężenie] $ \overline{a + \mi b} = a - b\mi $
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\item $ \alpha\overline{\alpha} = (a + b\mi)(a - b\mi) = a^2 + b^2 = |\alpha|^2 $
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\end{description}
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\subsection{Postać trygonometryczna}
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$ z = |z|\left( \frac{a}{|z|} + \frac{b}{|z|}\mi \right) $ ponieważ $ \sin\rho = \frac{b}{|z|} $ i $ \cos\rho = \frac{a}{|z|} $ mamy równość:
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$$ z = a + b\mi = |z|(\cos\rho + \mi\sin\rho) $$
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\begin{description}
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\item $ xy = |x|(\cos \alpha + \mi\sin\alpha) \times |y|(\cos \beta + \mi\sin\beta) =
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|x||y|\left[\cos(\alpha + \beta) + \mi\sin(\alpha+\beta)\right]$
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\item[Wzór de Moivre'a] $ z^n = |z|^n\left(\cos(n\rho) + \mi\sin(n\rho)\right) $
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\item[Pierwiastki] $ \sqrt[n]{ z } = \left\{ \sqrt[n]{|z|} \left(\cos \frac{\rho + 2k\pi}{n} + \mi\sin \frac{\rho + 2k\pi}{n} \right) \mid k = 0, 1, 2, ..., n-1 \right\} $
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\end{description}
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\end{document}
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