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If the product and the sum of two complex numbers are real, what can we say about the numbers? Prove it.

Explanation will be very helpful. Actually the question asks for a proof. Hence, a proof is all that we need. I actually am quite confused with it. I have proved a lot more complex questions in complex numbers and this little proof blocked my brain. Please stop this block.

Thanks in advance!

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  • $\begingroup$ Think about two general complex numbers, $a + bi$ and $c + di$. What is their product and their sum? What are necessary constraints on a, b, c, and d that allow for the product and the sum to be real? $\endgroup$ – Michael Stachowsky Sep 9 '16 at 16:09
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Let the two complex numbers be $z = x + iy$ and $w = u + iv$. Then, calculate $zw$ and $z+w$ and set the imaginary parts to zero.

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If $a$ and $b$ are your two numbers, they are the roots of the equation $$ (x-a)(x-b)=0 $$ which can also be written as $$ x^2-(a+b)x+ab=0 $$

The problem conditions mean that this is a quadratic equation with real coefficients. What does that tell you about $a$ and $b$?

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    $\begingroup$ I don't understand what you get if you establish this polynomial equation? $\endgroup$ – miracle173 Sep 9 '16 at 17:10
  • $\begingroup$ Roots of polynomials with real coefficients are either real or they come in pairs of conjugate numbers. $\endgroup$ – sometempname Sep 9 '16 at 17:16
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$$z_1=a+ib$$ $$z_2=c+id$$

We demand that $z_1+z_2$ and $z_1 \cdot z_2$ is real. If we assume that $z_1$ and $z_2$ are non-real, we have

$$a+ib+c+id=(a+c)+i(b+d) \implies b=-d.$$

$$(a+ib)(c+id)=(ac-bd)+i(ad+bc) \implies ad=-bc\implies a=c.$$

You can therefore write $$z_2=a-bi,$$

which is the complex conjugate of $z_1$.

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  • $\begingroup$ You should have that $a + ib + c + id = (a + c) + i\mathbf{(b+d)}$, not $(a + c) + i(b + c)$, and so we have that $b = -d$ instead of $b = -c$. $\endgroup$ – Dylan Sep 9 '16 at 16:38
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    $\begingroup$ @Dylan Thank you! Going in I also expected to get the conjugate.. I guess I trust my afternoon calculation skills way too much :) $\endgroup$ – Bobson Dugnutt Sep 9 '16 at 16:41
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    $\begingroup$ 0 and 1 sum/multiply to a real result, but are not complex conjugate... $\endgroup$ – Semiclassical Sep 9 '16 at 16:45
  • $\begingroup$ This answer is missing the general solution $b=d=0$ (with $a \ne c$). $\endgroup$ – TonyK Sep 9 '16 at 16:45
  • $\begingroup$ @TonyK Thank you for your comment. I've edited my answer accordingly, although I haven't calculated the case, as I suspect that was not the intention of the question. $\endgroup$ – Bobson Dugnutt Sep 9 '16 at 16:53
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Consider two complex numbers $q$ and $p$ such that,$q=c+id$ and $p=a+ib$. A.T.Q. $pq$ is real and $p+q$ real $$pq=(ac-bd)+(bc+ad)i$$ since pq is real, the imaginary part must be zero,hence bc+ad=0 or $$bc=-ad$$ then $$p+q=(a+c)+(b+d)i$$ here also $b+d$ must be zero or $$d=-b$$ So putting $d=-b$ in $bc=-ad$, you will get,$$b(c-a)=0$$ So either $$b=d=0$$ or $$a=c, b=-d$$.

So either $p$ and $q$ are conjugated $$p=a+bi,q=a-bi$$ or both are real $$p=a,q=c$$

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  • $\begingroup$ why a downvote?? $\endgroup$ – Vidyanshu Mishra Sep 9 '16 at 16:47
  • $\begingroup$ This answer is basically unreadable. Use Mathjax to express your answers symbolically, and space things properly. Also, 0 and 1 sum and multiply to real results but are not complex conjugates! $\endgroup$ – Semiclassical Sep 9 '16 at 16:49
  • $\begingroup$ Also: $z$ and $p$? Why?? $\endgroup$ – TonyK Sep 9 '16 at 16:50
  • $\begingroup$ p because i cant't write 1 and 2 in subscript. $\endgroup$ – Vidyanshu Mishra Sep 9 '16 at 16:51
  • $\begingroup$ but sorry guys i should have expressed it properly its my fault. $\endgroup$ – Vidyanshu Mishra Sep 9 '16 at 16:52
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you will have $$z_1+z_2=a,z_1\cdot z_2=b$$, setting $$z_1=x+iy,z_2=u+iv$$ then we get $$y+v=0$$ and $$yu+vx=0$$

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Let $z_1$ and $z_2$ be these copmlex numbers.

As their sum is real, we can write $z_1 = a + bi,\,z_2 = c - bi$ for $a, b, c \in \mathbb{R}$.

The product $z_1 z_2 = (a + bi) (c - bi) = ac + b^2 + b(c-a)i$ is real. Then $b(c-a) = 0$.

So $z_1$ and $z_2$ are conjugative (and, maybe, even real ones).

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    $\begingroup$ 0,1 aren't conjugate but satisfy the stated conditions. $\endgroup$ – Semiclassical Sep 9 '16 at 16:17
  • $\begingroup$ I've told they may just be real both $\endgroup$ – Denis Korzhenkov Sep 9 '16 at 16:18
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    $\begingroup$ If that's what you intended, the wording of that last line is confusing . What would be valid is: Either $z_1$ and $z_2$ are conjugate, or they are both real. $\endgroup$ – Semiclassical Sep 9 '16 at 16:20
  • $\begingroup$ thanks, @Semiclassical, that's right $\endgroup$ – Denis Korzhenkov Sep 9 '16 at 17:40
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If $r$ and $s$ are your complex numbers then the polynomial $(z-r)(z-s) = z^2-(r+s)z +rs$. The coefficients are the sum and product and must be real. So the polynomial is of the form $z^2+az+b$ with $a$ and $b$ real. If you set it equal to zero, and solve, the roots (which are $r$ and $s$) must be conjugates.

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    $\begingroup$ $0$ and $1$ are not conjugate, but they satisfy the stated conditions. $\endgroup$ – Micah Sep 9 '16 at 16:15
  • $\begingroup$ Um...yes they are. But I'll remove the word "complex". $\endgroup$ – B. Goddard Sep 9 '16 at 16:18
  • $\begingroup$ @B.Goddard Two complex numbers $z_1$ and $z_2$ are conjugate iff $z_1=a+bi$ and $z_2=a-bi$ for some real $a, b$. $0$ and $1$ definitely are not conjugate (note that the conjugate of a real number is just itself). $\endgroup$ – Noah Schweber Sep 9 '16 at 16:21
  • $\begingroup$ If you meant that the set of roots $\{z_1,z_2\}$ is preserved under complex conjugation, that's fine. I read it instead as $r$ and $s$ being conjugate to each other (which needn't be true). $\endgroup$ – Semiclassical Sep 9 '16 at 16:22
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    $\begingroup$ xkcd.com/169 $\endgroup$ – Micah Sep 9 '16 at 17:12

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