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I am trying to solve a complex exponential functional equation. A real to complex function $\gamma(t):\mathbb{R}\to\mathbb{C}$ has the following form \begin{align} \gamma(t)=e^{rt}\cdot e^{i(H(t)+2k\pi)} \end{align} where $r$ is a real constant and $k$ is any integer. $e^{rt}$ is the modulus of $\gamma(t)$ and $H(t)$ is the principle value of the argument of $\gamma(t)$. I know it satisfies the following functional equation \begin{align} [\gamma(t_1+t_2)]^2=[\gamma(t_1)\gamma(t_2)]^2 \end{align} So, I plug in $\gamma(t)$ to get \begin{align} [e^{r(t_1+t_2)}\cdot e^{i(H(t_1+t_2)+2k_1\pi)}]\cdot[e^{r(t_1+t_2)}\cdot e^{i(H(t_1+t_2)+2k_2\pi)}]=[e^{rt_1}\cdot e^{i(H(t_1)+2k_3\pi)}\cdot e^{rt_2}\cdot e^{i(H(t_2)+2k_4\pi)}]\cdot[e^{rt_1}\cdot e^{i(H(t_1)+2k_5\pi)}\cdot e^{rt_2}\cdot e^{i(H(t_2)+2k_6\pi)}] \end{align} The real part can cancel out so that \begin{align} e^{i2H(t_1+t_2)}\cdot e^{2k_1\pi}\cdot e^{2k_2\pi}=e^{i2H(t_1)}\cdot e^{i2H(t_2)}\cdot e^{2k_3\pi}\cdot e^{2k_4\pi}\cdot e^{2k_5\pi}\cdot e^{2k_6\pi} \end{align} This could be further simplified as \begin{align} &e^{i2H(t_1+t_2)}=e^{i2H(t_1)}\cdot e^{i2H(t_2)}\cdot e^{i2k\pi}\\ &e^{i2[H(t_1+t_2)-H(t_1)-H(t_2)]}=e^{i2k\pi} \end{align} Therefore, we have \begin{align} H(t_1+t_2)-H(t_1)-H(t_2)=k\pi \end{align} Adding $k\pi$ to both sides of equation, this is equivalent to \begin{align} H(t_1+t_2)+k\pi=H(t_1)+k\pi+H(t_2)+k\pi \end{align} Define function $f(t)\equiv H(t)+k\pi$ so that \begin{align} f(t_1+t_2)=f(t_1)+f(t_2) \end{align} which is the Cauchy functional equation, thus $f(t)=\lambda t$ where $\lambda$ is a real number. And $H(t)=\lambda t+k\pi$ where $k$ is an integer. Eventually, $\gamma(t)=e^{rt}\cdot e^{i\lambda t+k\pi}$. Since $e^{k\pi}=-1$ when $k$ is an odd number, we have $\gamma(t)=e^{rt}\cdot e^{i\lambda t}$ or $-e^{rt}\cdot e^{i\lambda t}$.


Is this the correct solution?

Also, I think that $[\gamma(t_1+t_2)]^2=[e^{r(t_1+t_2)}\cdot e^{i(H(t_1+t_2)+2k_1\pi)}]\cdot[e^{r(t_1+t_2)}\cdot e^{i(H(t_1+t_2)+2k_2\pi)}]$ but not $[e^{r(t_1+t_2)}\cdot e^{i(H(t_1+t_2)+2k\pi)}]^2$, is this correct? The former one have different $k_1$ and $k_2$ for the square of $\gamma(t_1+t_2)$, while the latter one has only one $k$. I feel it is very tempted to write as the latter one.

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  • $\begingroup$ Is $k$ an integer? $\endgroup$
    – saulspatz
    Sep 10 '19 at 18:14
  • $\begingroup$ Yes. $k$ is any integer $\endgroup$
    – zxjroger
    Sep 10 '19 at 18:17
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    $\begingroup$ Then you don't need to carry it at all. $e^{2k\pi i}=1$ $\endgroup$
    – saulspatz
    Sep 10 '19 at 18:20
  • $\begingroup$ Your functional equation simplifies to $\gamma(t_1+t_2)=\pm\gamma(t_1)\gamma(t_2)$. $\endgroup$
    – mr_e_man
    Sep 10 '19 at 18:47
  • $\begingroup$ @saulspatz Thanks. Do you think it is legitimate to plug $\gamma(t)$ into the functional equation? I always have concern that since argument of complex number is a multi-valued function, it may cause some problem after plugging in. For example, $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$ is a set equality where $z_1$ and $z_2$ are complex numbers. Do I need to worry about this? $\endgroup$
    – zxjroger
    Sep 10 '19 at 18:56
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For $\gamma : \Bbb{R \to C^*} $ continuous satisfying $$(\gamma(t_1+t_2))^2=(\gamma(t_1)\gamma(t_2))^2 $$ $\log$ is an isomorphism $\Bbb{C^* \to C/2i\pi Z}$ thus applying it to both side $$2\log \gamma(t_1+t_2) =2 \log \gamma (t_1) + 2\log \gamma(t_2) \in \Bbb{C/2i\pi Z}$$ From a solution we can lift $\log \gamma : \Bbb{R \to C/2i\pi Z}$ to a continuous function $\log \gamma : \Bbb{R \to C}$ whose reduction $\bmod 2i\pi$ agrees,

we'll have for some function $N : \Bbb{R\to Z}$ $$2\log \gamma(t_1+t_2) =2 \log \gamma (t_1) + 2\log \gamma(t_2)+2i \pi N(t)\in \Bbb{C}$$ The continuity implies $N(t)=n$ is constant which means $$\log \gamma(t) = ts+i\pi n, \qquad \gamma(t) = e^{ts} (-1)^n$$

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  • $\begingroup$ Thank you for your help. I am not in this field so I need sometime to understand your proof. Two quick questions. 1, what does C* mean? 2, how should I know the value of n? Because I thought the solution should be either $e^{ts}$ or $-e^{ts}$, but since you claimed that n is constant, there seems to be only one solution. $\endgroup$
    – zxjroger
    Sep 11 '19 at 13:14
  • $\begingroup$ Do you know any textbook on how to treat the multi-valued argument function of complex number in functional equation? I think this is the major issue I have. Thanks. $\endgroup$
    – zxjroger
    Sep 11 '19 at 15:17

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