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    Complex arithmetic and the elementary functions

    For \(z\in\CC\), \(\cos z=\frac{e^{iz}+e^{-iz}}{2}\) and \(\sin z=\frac{e^{iz}-e^{-iz}}{2i}\), and \(\cosh z=\cos(iz)\) and \(\sinh z=-i\sin(iz)\). No continuous branch of \(\log z\) exists on \(\CC\setminus\{0\}\). On the slit plane \(\CC\setminus(-\infty,0]\) the principal branch \(\Log z=\ln\abs{z}+i\Arg z\), with \(\Arg z\in(-\pi,\pi)\), is holomorphic, and its limits from the two sides of the negative real axis differ by \(2\pi i\). For \(a\in\CC\) the power \(z^a=e^{a\log z}\) depends on the branch of \(\log z\): changing the branch multiplies \(z^a\) by \(e^{2\pi ika}\) for some \(k\in\ZZ\).

    Exponential definitions of $\cosh$ and $\sinh$

    For \(z\in\CC\), the hyperbolic cosine and hyperbolic sine are \[ \begin{aligned} \cosh(z)&\coloneqq\frac{e^{z}+e^{-z}}{2},\\ \sinh(z)&\coloneqq\frac{e^{z}-e^{-z}}{2}. \end{aligned} \]

    With \(iz\) in place of \(z\), the same expressions give the trigonometric functions: \(\cos(z)=\frac{e^{iz}+e^{-iz}}{2}=\cosh(iz)\) and \(\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=-i\sinh(iz)\).

    How are $\cosh$ and $\sinh$ written as exponentials?

    $\cos z = \cosh(iz)$

    Circular cosine is hyperbolic cosine on the imaginary axis: $\cos z=\cosh(iz)$.Suggest a slogan

    For every \(z\in\CC\), \[ \cos(z) = \cosh(iz). \]

    How is $\cos(z)$ written in terms of $\cosh$?

    $\sin z = \frac{1}{i}\sinh(iz)$

    Circular and hyperbolic sine differ by the imaginary change of variable: $\sin z=\sinh(iz)/i$.Suggest a slogan

    For every \(z\in\CC\), \[ \sin(z) = {1\over i} \sinh(iz). \]

    How is $\sin(z)$ written in terms of $\sinh$?

    $\cosh(x+iy)$ in terms of real and imaginary parts

    $\cosh(x+iy)$ splits as $\cosh x\cos y+i\sinh x\sin y$.Suggest a slogan

    For \(x,y\in\RR\), \[ \cosh (x+i y) =\cosh (x) \cos (y)+i \sinh (x) \sin (y). \]

    What is the angle addition formula for $\cosh(x+iy)$?

    $\sinh(x+iy)$ in terms of real and imaginary parts

    $\sinh(x+iy)$ splits as $\sinh x\cos y+i\cosh x\sin y$.Suggest a slogan

    For \(x,y\in\RR\), \[ \sinh (x+i y) =\sinh (x) \cos (y)+i \cosh (x) \sin (y). \]

    What is the angle addition formula for $\sinh(x+iy)$?

    Taylor series of $\cosh z$

    $\cosh z$ keeps exactly the even terms of the exponential series.Suggest a slogan

    For every \(z\in\CC\), \[ \cosh z=1+\frac{z^{2}}{2 !}+\frac{z^{4}}{4 !}+\frac{z^{6}}{6 !}+\cdots=\sum_{n=0}^{\infty} \frac{z^{2 n}}{(2 n) !}. \]

    What is the series expansion of $\cosh(z)$?

    Taylor series of $\sinh z$

    $\sinh z$ keeps exactly the odd terms of the exponential series.Suggest a slogan

    For every \(z\in\CC\), \[ \sinh z=z+\frac{z^{3}}{3 !}+\frac{z^{5}}{5 !}+\frac{z^{7}}{7 !}+\cdots=\sum_{n=0}^{\infty} \frac{z^{2 n+1}}{(2 n+1) !}. \]

    What is the series expansion of $\sinh(z)$?

    $\sin\theta$ and $d\theta$ under the substitution $z=e^{i\theta}$

    On the unit circle, replace trigonometry by rational functions: $\sin\theta=(z-z^{-1})/(2i)$ and $d\theta=dz/(iz)$.Suggest a slogan

    For \(\theta\in\RR\) and \(z=e^{i\theta}\) on the unit circle, \[ \begin{aligned} \sin(\theta) &= \frac{e^{i\theta} - e^{-i\theta}}{2i} = \frac{z - \inverseof{z}}{2i}, \\ d\theta &= \frac{dz}{iz}. \end{aligned} \]

    How is $\sin(\theta)$ written in $z$ for a contour integral on the unit circle, and what is $d\theta$?

    Rational parametrization of the unit circle

    The unit circle minus $-1$ is rationally parametrized by $(1+ix)/(1-ix)$, equivalently $e^{2i\arctan x}$.Suggest a slogan

    The map \[ x\mapsto\frac{1-x^2}{1+x^2}+i\,\frac{2x}{1+x^2}=\frac{1+ix}{1-ix},\qquad x\in\RR, \] is a bijection from \(\RR\) onto \(S^1\setminus\{-1\}\). Substituting \(x=\tan t\) for \(t\in\qty{-\frac{\pi}{2},\frac{\pi}{2}}\) gives \[ F(t)\coloneqq\frac{1-x^2}{1+x^2}+i\,\frac{2x}{1+x^2}=\cos 2t+i\sin 2t=e^{2it}. \]

    Since \((1+ix)(1-ix)=1+x^2\), we have \(\frac{1+ix}{1-ix}=\frac{(1+ix)^2}{1+x^2}=\frac{1-x^2+2ix}{1+x^2}\), and its modulus is \(\abs{1+ix}/\abs{1-ix}=1\). For \(t\in(-\pi/2,\pi/2)\) and \(x=\tan t\), \(1+x^2=\sec^2t\), so \(\frac{1-x^2}{1+x^2}=\cos^2t-\sin^2t=\cos2t\) and \(\frac{2x}{1+x^2}=2\sin t\cos t=\sin2t\). As \(t\) runs over \((-\pi/2,\pi/2)\), \(x=\tan t\) runs bijectively over \(\RR\) and \(2t\) runs over \((-\pi,\pi)\), so \(e^{2it}\) runs bijectively over \(S^1\setminus\{-1\}\).

    What is the standard rational parameterization of the circle in $\CC$?

    Principal value of the complex logarithm

    For \(z\in\CC\setminus\{0\}\), let \(\Arg(z)\in(-\pi,\pi]\) be the principal argument of \(z\), the unique \(\theta\in(-\pi,\pi]\) with \(z=\abs{z}e^{i\theta}\). The principal value of the logarithm is \[ \log(z)\coloneqq\ln\abs{z}+i\Arg(z),\qquad z\in\CC\setminus\{0\}, \] where \(\ln\) is the real natural logarithm.

    In polar form \(z=re^{i\theta}\) with \(r>0\) and \(\theta\in(-\pi,\pi]\), this reads \(\log(z)=\ln r+i\theta\). It satisfies \(e^{\log(z)}=z\), restricts to the principal branch \(\Log\) on \(\CC\setminus(-\infty,0]\), and is discontinuous at every point of \((-\infty,0)\), where \(\Arg\) jumps from values near \(-\pi\) to \(\pi\).

    Principal branch of the logarithm and of $z^\alpha$

    Let \(\Omega\coloneqq\CC\setminus(-\infty,0]\). Every \(z\in\Omega\) can be written uniquely as \(z=re^{i\theta}\) with \(r>0\) and \(\theta\in(-\pi,\pi)\). The principal branch of the logarithm is the function \(\Log\colon\Omega\to\CC\) given by \[ \Log(z)\coloneqq\log r+i\theta, \] where \(\log r\) is the real natural logarithm. For \(\alpha\in\CC\), the principal branch of \(z^\alpha\) is the function on \(\Omega\) given by \[ z^{\alpha}\coloneqq e^{\alpha\Log(z)}. \]

    Complex powers $z^\alpha$ along a branch of the logarithm

    Let \(\Omega\subseteq\CC\setminus\{0\}\) be open. A branch of the logarithm on \(\Omega\) is a continuous function \(L\colon\Omega\to\CC\) with \(e^{L(z)}=z\) for all \(z\in\Omega\). For such \(L\) and \(\alpha\in\CC\), the branch of \(z^\alpha\) determined by \(L\) is \[ z^\alpha\coloneqq e^{\alpha L(z)},\qquad z\in\Omega. \]

    When no branch is specified, \(L\) is the principal branch \(\Log\) on \(\CC\setminus(-\infty,0]\).

    Existence of a branch of the logarithm on a simply connected region

    On a simply connected region avoiding $0$, the complex logarithm has a single-valued holomorphic branch.Suggest a slogan

    Let \(\Omega\subseteq\CC\) be a simply connected region with \(1\in\Omega\) and \(0\notin\Omega\). Then there exists a holomorphic function \(F\colon\Omega\to\CC\) such that

    • \(e^{F(z)}=z\) for all \(z\in\Omega\), and

    • \(F(r)=\log r\) for all real \(r\) in some neighborhood of \(1\) in \(\Omega\).

    Holomorphic logarithm of a nonvanishing function on a simply connected domain

    On a simply connected domain, every nowhere-vanishing holomorphic function has a holomorphic logarithm.Suggest a slogan

    Let \(\Omega\subseteq\CC\) be a simply connected open set and let \(f\) be holomorphic and nonvanishing on \(\Omega\). Then there is a holomorphic function \(G\) on \(\Omega\) such that \[ f(z)=e^{G(z)}\quad\text{for all } z\in\Omega. \]