# How to solve a complex exponential functional equation which contains the multivalued argument function of complex numbers

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.

• Is $k$ an integer? Sep 10 '19 at 18:14
• Yes. $k$ is any integer Sep 10 '19 at 18:17
• Then you don't need to carry it at all. $e^{2k\pi i}=1$ Sep 10 '19 at 18:20
• Your functional equation simplifies to $\gamma(t_1+t_2)=\pm\gamma(t_1)\gamma(t_2)$. Sep 10 '19 at 18:47
• @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? Sep 10 '19 at 18:56

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 $$\log \gamma(t) = ts+i\pi n, \qquad \gamma(t) = e^{ts} (-1)^n$$
• 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. Sep 11 '19 at 13:14