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.
Thanks in advance!
 A: 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.
A: 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 $b$?
A: $$z_1=a+ib$$
$$z_2=c+id$$
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 $z_1$.
A: 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
$$p=a,q=c$$
A: 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 $$yu+vx=0$$
A: Seems no answer was accepted yet. Try this luck...
The product of two reals can be expressed by the complex quantities by
\begin{align}
\Re\{z_1\}\Re\{z_2\}&=\frac{1}{2}\Re\{z_1\}(z_2+z_2^\ast)=\frac{1}{2}\Re\{z_1z_2\}+\frac{1}{2}\Re\{z_1z_2^\ast\}    ..........(1)
\end{align}
Let's show the relationship (1) with any arbitrary  (complex) quantities $z_1=Ae^{i\omega_it}$ and $z_2=Be^{i\omega_jt}$, then
\begin{align}
\Re{\{Ae^{i\omega_it}\}}\Re{\{Be^{i\omega_jt}\}}&=\frac{1}{4}{\left\{(Ae^{i\omega_it}+A^*e^{-i\omega_it})(Be^{i\omega_jt}+B^*e^{-i\omega_jt})\right\}}\nonumber\\
&=\frac{1}{4}{\left\{ABe^{i\omega_{ij}^+t}+A^*B^*e^{-i\omega_{ij}^+t}+AB^*e^{i\omega_{ij}^-t}+A^*Be^{-i\omega_{ij}^-t}\right\}}\nonumber\\
&=\frac{1}{4}{\left\{ABe^{i\omega_{ij}^+t}+(ABe^{i\omega_{ij}^+t})^*+AB^*e^{i\omega_{ij}^-t}+(AB^*e^{i\omega_{ij}^-t})^*\right\}}\nonumber\\
&=\frac{1}{2}\Re{\left\{ABe^{i\omega_{ij}^+t}+AB^*e^{i\omega_{ij}^-t}\right\}}\nonumber\\
&=\frac{1}{2}\Re{\left\{(Ae^{i\omega_{i}})(Be^{i\omega_{j}t}+B^*e^{-i\omega_{j}t})\right\}}\\
&=\frac{1}{2}\Re{\left\{Q_{ij}^+e^{i\omega_{ij}^+t}\right\}}+\frac{1}{2}\Re{\left\{Q_{ij}^-e^{i\omega_{ij}^-t}\right\}}
\end{align}
where $\omega_{ij}^+=\omega_i+\omega_j$ and $\omega_{ij}^-=\omega_i-\omega_j$. It can be seen that $AB=BA=Q_{ij}^+=Q_{ji}^+$ and $AB^*=(A^*B)^\ast=Q_{ij}^-=Q_{ji}^{-*}$
A: 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.  
A: 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).
