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

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.

• 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? – Michael Stachowsky Sep 9 '16 at 16:09

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.

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 $b$?

• I don't understand what you get if you establish this polynomial equation? – miracle173 Sep 9 '16 at 17:10
• Roots of polynomials with real coefficients are either real or they come in pairs of conjugate numbers. – sometempname Sep 9 '16 at 17:16

$$z_1=a+ib$$ $$z_2=c+id$$

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 $z_1$.

• 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 $b = -c$. – Dylan Sep 9 '16 at 16:38
• @Dylan Thank you! Going in I also expected to get the conjugate.. I guess I trust my afternoon calculation skills way too much :) – Bobson Dugnutt Sep 9 '16 at 16:41
• 0 and 1 sum/multiply to a real result, but are not complex conjugate... – Semiclassical Sep 9 '16 at 16:45
• This answer is missing the general solution $b=d=0$ (with $a \ne c$). – TonyK Sep 9 '16 at 16:45
• @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. – Bobson Dugnutt Sep 9 '16 at 16:53

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 $$p=a,q=c$$

• why a downvote?? – Vidyanshu Mishra Sep 9 '16 at 16:47
• 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! – Semiclassical Sep 9 '16 at 16:49
• Also: $z$ and $p$? Why?? – TonyK Sep 9 '16 at 16:50
• p because i cant't write 1 and 2 in subscript. – Vidyanshu Mishra Sep 9 '16 at 16:51
• but sorry guys i should have expressed it properly its my fault. – Vidyanshu Mishra Sep 9 '16 at 16:52

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 $$yu+vx=0$$

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).

• 0,1 aren't conjugate but satisfy the stated conditions. – Semiclassical Sep 9 '16 at 16:17
• I've told they may just be real both – Denis Korzhenkov Sep 9 '16 at 16:18
• 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. – Semiclassical Sep 9 '16 at 16:20
• thanks, @Semiclassical, that's right – Denis Korzhenkov Sep 9 '16 at 17:40

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.

• $0$ and $1$ are not conjugate, but they satisfy the stated conditions. – Micah Sep 9 '16 at 16:15
• Um...yes they are. But I'll remove the word "complex". – B. Goddard Sep 9 '16 at 16:18
• @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). – Noah Schweber Sep 9 '16 at 16:21
• 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). – Semiclassical Sep 9 '16 at 16:22
• xkcd.com/169 – Micah Sep 9 '16 at 17:12