$\begingroup$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!
$\endgroup$18 Answers
$\begingroup$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.
$\endgroup$$\begingroup$If $a$ and $b$ are your two numbers, they are the roots of the equation $$ (x-a)(x-b)=0 $$ which can also be written as $$ x^2-(a+b)x+ab=0 $$
The problem conditions mean that this is a quadratic equation with real coefficients. What does that tell you about $a$ and $b$?
$\endgroup$2$\begingroup$$$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,$$
which is the complex conjugate of $z_1$.
$\endgroup$5$\begingroup$Consider two complex numbers $q$ and $p$ such that,$q=c+id$ and $p=a+ib$. A.T.Q. $pq$ is real and $p+q$ real $$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$$
$\endgroup$12$\begingroup$you will have $$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$$
$\endgroup$$\begingroup$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 $AB^*=(A^*B)^\ast=Q_{ij}^-=Q_{ji}^{-*}$
$\endgroup$$\begingroup$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).
$\endgroup$4$\begingroup$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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