I understand that $\cos(z)=\cosh(iz)$ but I was wondering what happened at higher powers of $z$.
E.g does $\cos(z^2)=\cosh(iz^2)$ implying that $\cos(z^n)=\cosh(iz^n)$
Cos(z) and Cosh(iz)
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real-analysis
complex-analysis
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0Set $z=y^n$ in the original equation ... – 2017-02-27
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0Well, yes. That's how composition of functions (or call it "substitution principle" or "putting stuff into stuff") works. – 2017-02-27
1 Answers
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$\newcommand{\C}{\mathbb{C}}$It might be more clear if in the first equation we use a different variable besides $z$. I'll use $w$. We know $\cos w = \cosh(iw)$ for any and all complex numbers $w \in \C$. Hence if I have another complex number $z$, then $z^2, z^3, ..., z^n$ are also all complex numbers since raising a complex number to a positive integer yields another complex number. So all of these $z^n$'s are valid values for $w$ in our first equation and can be plugged in. Hence \begin{align*} \cos(z^n) = \cosh(iz^n) \end{align*}