Is there a pair of functions solving this "trigonometric-like" system of ODEs?

Question

The (ordinary) trigonometric functions are the solutions to the system of ordinary differential equations: \begin{align} c' &= -as, \\ s' &= ac, \end{align} with $c(0) = 1$ and $s(0) = 0$, for some constant $a$. Similarly, the hyperbolic functions are the solutions to the system \begin{align} c' &= as, \\ s' &= ac, \end{align} with the same initial conditions. Is there similarly some pair of functions satisfying \begin{align} c' &= \bar{z}s, \\ s' &= zc, \end{align} again with the same initial conditions, and where $z$ is complex and $\bar{z}$ is the complex conjugate of $z$?

I have tried combining the trigonometric and hyperbolic functions in various ways, but cannot find a solution. If such a pair of functions does not exist, is there a method of finding e.g. a series solution?

Answer 1

If we write $z = re^{i\theta}$, the solution is straightforwardly $$ c = \cosh rt, \\ s = e^{i\theta}\sinh rt, $$ which is easily generalized to non-constant $z$, only as long as the complex phase, $\theta$, is constant.

Answer 2

So, we start with the system

$c' = \bar zs, \tag 1$

$s' = z c, \tag 2$

with $z \in \Bbb C$ a constant. We differentiate (1) and obtain

$c'' = \bar z s' = \bar z z c, \tag 3$

where we have made use of (2); now assuming $z \ne 0$ we may write

$z = re^{i\theta}, \tag 4$

whence

$0 < \bar z z = r^2 \in \Bbb R; \tag 5$

thus (3) takes the form

$c'' = r^2c; \tag 6$

we are given that $c(0) = 1$ and $s(0) = 0$; from this latter condition we deduce immidiately, via (1), that $c'(0) = 0$ and now we have a complete set of initial conditions for the second-order equation (3); since $r^2 > 0$ we know that the solutions to (3) will be linear combinations of $e^{rt}$ and $e^{-rt}$ with $r = \sqrt{\bar z z} > 0$; that $c(0) = 1$ and $c(0) = 0$ implies that this linear combination is in fact

$c(t) = \cosh rt ; \tag{7}$

from (1) and (7) we immediately infer that

$s(t) = \dfrac{1}{\bar z} c'(t) = r^{-1}e^{i\theta} (\cosh rt)' = r^{-1}e^{i\theta}(r \sinh rt) = e^{i\theta} \sinh rt. \tag{8}$