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\chapter Euler's formula
\section Complex numbers \label sec:comlex

One can introduce \em{complex numbers} by considering so-called \em{imaginary unit} $i$, such that:
\equation \label eq:i
    i^2 = -1.
\definition
    \emph{Complex number} is a number of a form $x+iy$, where $x, y \in \mathbb R$.

\section Exponent of complex number \label sec:exp
\theorem \label thm:1
    Let $x$ be a real number. Then
    \equation \label eq:main
        e^{ix} = \cos x+i\sin x.
\proof
    Let us recall series for exponent, sine and cosine:
    \align
        \item \label eq:series-exp
            e^z &= 1 + z + \frac{z^2}{2} + \frac{z^3}{3!} + \ldots = \sum_{k=0}^\infty \frac{z^k}{k!}
        \item \label eq:series-sin
            \sin z &= z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \ldots
        \item \label eq:series-cos
            \cos z &= 1 - \frac{z^2}{2} +\frac{z^4}{4!} - \frac{z^6}{6!} + \ldots
    It follows from \ref{eq:i} that
    \eq
        i^k=\begin{cases}
            1 & \text{for } k = 4m\\\\
            i & \text{for } k = 4m+1\\\\
            -1 & \text{for } k = 4m+2\\\\
            -i & \text{for } k = 4m+3
            \end{cases}
    Let us put $z=ix$ into \ref{eq:series-exp}. For even $k$, $(ix)^k$ is real and for odd $k$ it is imaginary. Moreover, the sign alternates when one pass to the next term. It follows immediately that the real part of \ref{eq:series-exp} is equal to \ref{eq:series-sin} and the imaginary part is equal to \ref{eq:series-cos} with substitution $z=ix$.

This finished the proof of \ref[Euler's formula\nonumber][thm:1].

\chapter Corollary

It follows from \ref[Theorem][thm:1] from \ref[Section][sec:exp] that
\eq
    \sin x = \frac{e^{ix}-e^{-ix}}{2i}.
Therefore,
\align
    \item \nonumber
        \sin 2x &= \frac{e^{2ix}-e^{-2ix}}{2i} = \frac{1}{2i}((e^{ix})^2-(e^{-ix})^2)=
    \item \nonumber
        &\frac{1}{2i}(e^{ix}-e^{-ix})(e^{ix}+e^{-ix})=2\sin x \cos x.

\question
    Express $\cos x$ in terms of exponents. (Click on pencil to check the correct answer.)
    \quiz
        \choice
            $(e^{ix}+e^{-ix})/(2i)$
            \comment
                No, this is complex number!
        \choice \correct
            $(e^{ix}+e^{-ix})/2$
            \comment
                Yes! That's right!
        \choice
            $(e^{ix}-e^{-ix}))/(2)$
            \comment
                No, that's $-i\sin x$.

\exercise
    Express $\cos 2x$ in terms of $\sin x$ and $\cos x$.