Out of curiosity, does anyone happen to know if it's possible to write a complex number $z\in\mathbb{C}$ in terms of a series in trig functions ($\sin$, $\cos$, $\tan$, $\sec$, $\csc$, $\cot$, etc.) and/or hyperbolic trig functions ($\sinh$, $\cosh$, $\tanh$, sech, csch, $\coth$, etc.)? Ie, something of the form $$z=\sum_{n=1}^\infty a_n\text{trig}_n(b_nz),$$ where each $\text{trig}_n$ is a trig or hyperbolic trig function (which doesn't necessarily have to be the same for each $n$) and $a_n$,$b_n$ are (possibly compex) constants (independent of $z$)?

  • $\begingroup$ Sounds like Fourier series might be a topic worth looking into since they can express "nice enough" functions as a (possibly infinite) sum of sine and/or cosine functions $\endgroup$ – Eevee Trainer May 13 at 21:57
  • $\begingroup$ But I think that really only works for periodic functions, and $z$ is not periodic. $\endgroup$ – arow257 May 13 at 22:06

Choosing a definite argument $\;t\in[0,2\pi)\;$ for $\;z\in\Bbb C\;$ , we can write

$$z=|z|e^{it}=|z|\cos t+i|z|\sin t$$

There, you wrote $\;z\in\Bbb C\;$ as a sum of trigonomeric functions (multiplied each by some constant). Certainly, the case $\;z=0\;$ should be dealt with separatedly...

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  • $\begingroup$ But I wanted all the $z$ dependence inside the trig functions. The problem is trivial otherwise. $\endgroup$ – arow257 May 13 at 22:08
  • $\begingroup$ @arow257 I've no idea what may "all z dependence inside trig function" possibly mean...sorry. And yes: it is a rather trivial problem as you posed it. $\endgroup$ – DonAntonio May 13 at 22:17
  • $\begingroup$ It means something of the form $a_n\cdot \text{trig}(b_n z)$ with $a_n$ and $b_n$ (possibly complex) constants. I will edit my question to clarify. $\endgroup$ – arow257 May 13 at 22:20
  • $\begingroup$ ...and $$z=|z|\cos\left(\arg z\right)+i|z|\sin\left(\arg z\right)\ldots?$$ Here, $$a_1=|z|,\,b_1=\arg z\,,\,\,a_2=i|z|\,,\,\,b_2=\arg z$$ and, of course, $\;a_n=b_n=0\;$ for $\;n\ge3\;$ ...This, of course, is exactly the same that in my answer but written in a much more cumbersome way. If you'd add some context to your question then it'd be probably easier to know what are you expecting... $\endgroup$ – DonAntonio May 13 at 23:08
  • $\begingroup$ But your $a_1$, $a_2$ depend on $z$... I asked for $a_n$ and $b_n$ to be constants, i.e., independent of $z$. $\endgroup$ – arow257 May 13 at 23:16

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