wavefunction - Schrödinger's Equation and its complex conjugate

I would like to know why there is a minus sign on the right-hand side of the Schrödinger's complex conjugate equation, whereas in the Schrödinger's equation there isn't. I know it is a simple question, but I don't know where this comes from.

$$-\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi = i \hbar \frac{\partial \psi}{\partial t}$$

$$-\frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2} + V(x)\psi^* = -i \hbar \frac{\partial \psi^*}{\partial t}$$

Answer

It is the definition of complex number. Let's say

$z=x+iy\quad \Rightarrow z^*=x-iy$

$z=x-iy\quad \Rightarrow z^*=x+iy$

In simple words, you just have to change the sign of the Imaginary part. The thing is that $\psi(x)$ it's a imaginary number, so it's conjugate it's just $\psi^*(x)$. If you have the $\psi(x)$ function, then you can change $i\to -i$ or in the oposite way.