Finding the solutions of an equation of the form $f(z)=0$ That's how I understand it: usually, for finding the solutions of an equation $f(z)=0$, we try to convert this expression to a "simpler one" from which we can deduce
which values can satisfy it; thus at the end of the procedure we arrive to an implication of the form
$$f(z)=0 \Rightarrow z=k , k \in S,$$
but what a number $k$ need to satisfy in order to call it a solution for the equation $f(z)=0$ is the other implication $$z=k \Rightarrow f(z)=0.$$ Then, we should not rush and call those numbers $k$ solutions, but possible solutions of the equation.
There isn't a big deal when working with simple equations like a quadratic $ax^2+bx+c=0$; by the Fundamental Theorem of Algebra, it has two (maybe equal to each other)
and only two solutions. Since the quadratic formula yields one or two numbers, we are sure those are the solutions of the equation.
But now that I'm starting learning Complex Analysis, I need to deal with weirder equations like
$$cos(z)=4$$ (Exercise 3-b) page 38, Basic Complex Analysis, Marsden).
After a few calculations I arrived to the set of solutions
$$z=2 \pi n +i(-log(4+sqrt(15))),$$ $$z=2 \pi n +i(-log(4-sqrt(15))),$$ with $n \in \mathbb{Z}$.
My question is: Am I right thinking that once we arrived to the numbers above, we need to check that those are actually solutions of the original equation?
If so, could someone please give me an example of an equation from which one of the possible solutions turns out not to be a solution of the equation?
Thank you in advance for your comments.
 A: If, while doing your computations, you only del with equivalences, there is nothing to check. An example would be:$$(e^z)^2=1\iff e^{2z}=1\iff 2z=2k\pi i\ (k\in\Bbb Z)\iff z=k\pi i\ (k\in\Bbb Z).$$
But if you also have implications which are not equivalences, then, yes, you have to check whether or not all the numbers that you got are indeed solutions. Note this happens in general, not only when you deal with complex numbers.
A: You are right. It's a common mistake that high school teachers make when they teach to solve even the most simple equations.
As José Carlos Santos says in his answer, you can track which of the steps of a solution are only implications or are actually equivalences.
Here's an easy example where not all the obtained possible solutions are solutions:
Take \begin{align}
& x = \sqrt{2-x^2}\\
\implies \quad & x^2 = 2-x^2\\
\implies \quad & 2x^2 = 2\\
\implies \quad & x^2 = 1\\
\implies \quad & x = 1 \text{ or } x=-1
\end{align}
But $-1$ is obviously not a solution.
For a slightly more interesting example try to solve (in $\Bbb C$) the equation $$\sqrt{1+z}+\sqrt{1-z} = 1$$
