Differential equation in Complex numbers How do I solve this differential equation
$$Z'=Z^*$$
such that $Z$ is complex numbers and $Z^*$ is the conjugate of $Z$?
Sketch the phase portraits of solutions.
 A: Writing
$z = x + iy, \tag 1$
where $x$, $y$ and $z$ are functions of the independent variable $t$, we have
$\dot z = \dot x + i \dot y, \tag 2$
and
$z^\ast = x - iy; \tag 3$
then the given equation
$\dot z = z^\ast \tag 4$
becomes
$\dot x + i \dot y = x - iy; \tag 5$
equating the real and imaginary parts we find
$\dot x = x \tag 6$
and
$\dot y = -y; \tag 7$
to proceed further, we assign $z$ the initial condition
$z(t_0) = x(t_0) + iy(t_0), \tag 8$
which in fact initializes $x$ and $y$; now (6) and (7) have the well-known unique solutions
$x(t)= x(t_0)e^{t - t_0}, \tag{9}$
$y(t) = y(t_0)e^{t_0 - t}, \tag{10}$
whence
$z(t)  = x(t_0)e^{t - t_0} + iy(t_0)e^{t_0 - t}. \tag{11}$
In order to sketch the phase portrait of the system (4), we first take note of the fact that (9) and (10) together yield
$x(t)y(t) = x(t_0)y(t_0), \tag{12}$
which whenever
$x(t_0) \ne 0 \ne y(t_0) \tag{13}$
shows that the solution curves lie in he hyperbolas
$xy = x(t_0)y(t_0) \tag{14}$
in the $xy$-plane $\Bbb R^2$; if precisely one of $x(t_0)$, $y(t_0)$ does not vanish, then the corresponding phase curve is either the positive or negative $x$ or $y$ half-axis.  For example, if
$x(t_0) > 0, \; y(t_0) = 0, \tag{15}$
the solution curve traverses the entire positive $x$-axis as $-\infty \to t \to \infty$.  Similar results apply when
$x(t_0) = 0, \; y(t_0) \ne 0 \tag{16}$
etc; the reader may easily complete the necessary details; finally when
$x(t_0) = 0 = y(t_0) \tag{17}$
the phase curve is the single point $(0, 0)$.  With these facts in mind, the phase portrait of (4) may be sketched with little additional effort.
