TITLE: Proving complex numbers
QUESTION [6 upvotes]: Let $z_1,z_2$ be two complex numbers such that $z_1 + z_2$ and
  $z_1\dot\ z_2$ are each negative real numbers. Prove that $z_1$ and
  $z_2$ must be real numbers.

My attempt at a solution follows,
Let $z_1 = a+bi$ and $z_2=c+di$. The hypotheses states that $a+c<0$, $b+d = 0$, $ac-bd <0$, and $ad + bc=0$. And I get stuck here since I do not know the logic to proceed in which I can prove that $z_1$ and $z_2$ must be real numbers.

REPLY [2 votes]: To add something none of the other answers have added explicitly....
You have chosen to splitting the complex numbers into real and imaginary parts. To that end, you have faithfully translated the givens of your problem into this form.
What you've overlooked is that you haven't translated your goal into this form. Doing this isn't always the simplest way to proceed, but it's almost always a correct and straightforward way to proceed. The translation of your goal is that you seek to prove $b=d=0$.
Finally, even without an idea of how to proceed, you do know something you can do with a system of equations and inequations: you can simply / solve it! Sometimes, the simplified/solved form will suggest what to do next. So without knowing where you're going, we can still do the following:

Solve $b+d = 0$ to determine $d = -b$
Simplify $ad+bc = 0$ to $b(c-a) = 0$
Solve that equation by splitting into cases:
Case $b=0$:

Then $d=0$ too.
The remaining inequalities are $a+c < 0$ and $ac - bd < 0$
The latter simplifies to $ac < 0$
We can solve this by cases:
Case $a < 0$ and $c > 0$:

The remaining inequation yields $a < -c < 0$

Case $a > 0$ and $c < 0$:

The remaining inequation yields $0 < a < -c$


Case $(c-a)=0$:

Then $c = a$
The remaining inequalities are $a+c < 0$ and $ac - bd < 0$
Thes latter simplifies to $a^2 < -b^2$
This is a contradiction! (the r.h.s is nonpositive and the l.h.s. is nonnegative)


So the complete solution to the givens is the union of the cases:

$a < 0$ and $c > 0$ and $|c| < |a|$ and $b=d=0$
$a > 0$ and $c < 0$ and $|a| < |c|$ and $b=d=0$

And then once you've reached this point, you could return to the problem to see how you could use this information to show $z_1$ and $z_2$ are real numbers.
(Of course, if you translated the goal, you wouldn't have had to do any of the work in the $b=0$ case)