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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.

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$d=-b\Rightarrow d(a-c)=0$ in your last equality $\to$ either $d$ or $a-c$ is $0$. So... – L. F. Jan 17 '13 at 1:21
@LordoftheFlies I am not sure what you are trying to do? So in order to prove that they are real I mist show d or a is 0? – Q.matin Jan 17 '13 at 1:24
$a-c<0$ so it cannot be $0$. Therefore $d=0$ i.e. $z_2$ is real. Since $d=-b\Rightarrow b=0$ i.e. $z_1$ is real. – L. F. Jan 17 '13 at 1:26
up vote 6 down vote accepted

From $b+d=0$, get $d=-b$. Then from $ad+bc=0$ get $b(c-a)=0$.

So either $b=0$, from which $d=0$ making $z_1,z_2$ real, or else $c=a$ (and $d=-b$ still holds). Now put those into $ac-bd<0$ and it becomes $a^2+b^2<0$, contradiction.

NOTE: Give credit to "Lord of the Flies", whose comment came in while I was writing this.

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Thanks but then what follows directly that says $z_1$ and $z_2$ are real numbers? – Q.matin Jan 17 '13 at 1:34
The argument above shows that either both $b$ and $d$ are zero, or else a contradiction occurs. Can conclude that is must be that $b=d=0$. The $b,d$ are the imaginary parts of the two complex numbers $z_1,z_2$ using your notation, and saying imaginary part is $0$ is the same as saying a complex number is real. – coffeemath Jan 17 '13 at 1:36
Ahh, a bulb just turned on in my head when you said, "saying imaginary part is 0 is the same as saying a complex number is real". Thanks a lot coffeemath!! – Q.matin Jan 17 '13 at 1:41

$b+d=0 \implies d=-b \implies ad+bc=-ab + bc = 0\implies b(c-a)=0$.

  • So either $b=0$, from which $d= b = 0$, and thus $z_1,z_2$ are real,
  • or else $c-a= a-c = 0$ which contradicts the fact that $a - c < 0.$

Therefore $z_1, z_2 \in \mathbb{R}$.

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Thanks but did you mean $c-a$ instead of $a-c$? – Q.matin Jan 17 '13 at 1:36
amWhy: Is $a-c<0$ a hypothesis, or is it $a+c<0$? – coffeemath Jan 17 '13 at 1:38

Alternately, since $z_1+z_2$ is real and $z_1z_2$ is negative real, and $$(z_1-z_2)^2=(z_1+z_2)^2-4z_1z_2,$$ we conclude that $(z_1-z_2)^2$ is positive. Thus $z_1-z_2$ is real. Since $z_1+z_2$ is real, it follows by addition that $2z_1$, and therefore $z_1$, is real.

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Isn't $z_1+z_2$ also negative real? – Q.matin Jan 17 '13 at 1:51
Yes it is negative real, but it gets squared on the right side so becomes positive in its contribution. – coffeemath Jan 17 '13 at 1:54
In the problem, it is. But the above proof does not need or use the fact that $z_1+z_2$ is negative. It does use the fact that the product $z_1z_2$ is negative. – André Nicolas Jan 17 '13 at 1:54

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)

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Thank so much!! THis is very helpful! – Q.matin Jan 18 '13 at 1:46

$ (x\!-\!z_1)(x\!-\!z_2) = x^2\! - (z_1\!+\!z_2) x + z_1 z_2\in \Bbb R[x]$ has discriminant $\,(z_1\!+\!z_2)^2\!-4z_1z_2 > 0,$ since $z_1z_2 < 0,$ hence the roots $z_1,z_2$ are real (by the quadratic formula).

Remark $\ $ The proof works more generally for $\ z_1,z_2$ elements of any (integral) domain $\,\Bbb D\supset \Bbb R,$ assuming only that $\,z_1\!+z_2,\, z_1 z_2\,$ are both in $\Bbb R,$ and $\,z_1 z_2 < 0$. The hypothesis that $\Bbb D $ is a domain ensures that the list of roots $z_1,z_2 \in \Bbb R$ obtained by the quadratic formula persists as the unique list of roots in $\Bbb D$ (if the quadratic polynomial had an additional root $z_3\in \Bbb D$ then it would have more roots than its degree, which cannot occur in a domain). The claim may fail in extension rings not domains, e.g. $ f(x) = (x+2)(x-1) = x^2+x-2 = (x-w)(x+w+1)\, $ over $\,\Bbb R[w]/(w^2+w-2),\,$ where $f$ has non-real roots $\ w,\, -w-1 \not\in \Bbb R,\,$ contra to the claim.

Notice that said extension ring is not a domain since there $\,(w+2)(w-1) = 0,\,$ but $\,w \ne -2,1$. Informally, one may think of $w$ as an algebraic representation of a "generic" root of $f$, sharing all $(\Bbb R -)$algebraic properties of the roots $-2,1,$ i.e. $w$ satisfies $f(w) = g(w)$ for polynomials $f,g\in\Bbb R[x]\,$ iff $\,-2$ and $1$ also do: $f(-2) = g(-2)$ and $f(1) = g(1).\,$ While we can always construct extension rings containing such generic roots, generally the construction does not preserve the property of being a domain, and the consequent uniqueness of the list/multiset of roots (which, suitably formulated, is a characteristic property of domains among rings). Yet another example that uniqueness theorems provide powerful tools for deducing equalities.

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...from which we also see that the hypotheses can be weakened to "$z_1+z_2 \in \mathbb{R}$ and $z_1 z_2 < 0$". – Hans Lundmark Jan 17 '13 at 10:13
@Hans Yes, in fact it works much more generally - see my recently added remark. – Math Gems Jan 17 '13 at 17:19
@MathGems Thanks a lot! – Q.matin Jan 18 '13 at 1:46

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