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Let $z, w \in \mathbb{C}$. Prove that if $zw$ and $z + w \in \mathbb{R}$, then $z = \bar{w}$.So I have started to see complex numbers in class and by reading I find this problem. So I know that $$Re(z)= \frac{z + \bar{z}}{2},$$ but I don't seem to get to the answer. I also tried using other basic properties of $\mathbb{C}$.

Any help/hints?

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  • $\begingroup$ Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. $\endgroup$
    – Martin R
    May 26, 2020 at 7:15

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Let $z=a+ib$, $\omega = c + id$, with $a,b,c,d \in \mathbb{R}$. We know that: $$ z \omega = ac + iad + ibc - bd \in \mathbb{R} $$ This implies that: $$ ad+bc=0 $$ that is, $ad=-bc$. Moreover, we also know that: $$ z + \omega = a+c + i(b+d) \in \mathbb{R}$$ which implies that $$ -b=d $$ Consequently: $$ ad= dc $$ Suppose $d \neq 0$. Then, this implies that $a=c$, and thus: $$ z= a +ib = c - id = \overline{\omega} $$ If $d=0$, then $\omega = c$ and: $$ b=-d=0 $$ that is, $z=a$. It is then clear that the result does not hold in this case: indeed, if this were true, it would mean that just because the sum of two real numbers is real, and their product is real, the number are certainly equal, which is absurd.

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The quadratic equation $t^2 - (z+u) t + zu =0$ has solutions $z$ and $u$.

Since $z+u, zu \in \mathbb{R}$, we have $\overline{t^2 - (z+u) t + zu} = \overline{t}^2 - (z+u) \overline{t} + zu = \overline{0} = 0$, which implies that if $t$ is a solution then $\overline{t}$ also is. This lead us to $z = \overline{u}$.

You can prove much general thing: If $z$ is a zero of some real-coeffienct polynomial $p(t)$, $\overline{z}$ is another zero of $p$.

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    $\begingroup$ This is the most elegant answer! $\endgroup$
    – rae306
    May 26, 2020 at 9:35
  • $\begingroup$ @rae306 hey, have you seen my answer? $\endgroup$ May 26, 2020 at 19:30
  • $\begingroup$ @AnasA.Ibrahim i havn’t $\endgroup$
    – dust05
    May 26, 2020 at 22:10
  • $\begingroup$ @dust05 It is written here in the answers, you can read it. It's not a very rigorous proof, but I don't think there's a mistake in it. $\endgroup$ May 26, 2020 at 23:24
  • $\begingroup$ @AnasA.Ibrahim i don’t see any mistake neither. I think the notion of complex plane and modulus-angle thing are unfamiliar to somebody. $\endgroup$
    – dust05
    May 27, 2020 at 0:57
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Write $z=x+iy$ and $w=r+is$. We know that $z+w$ is real and this means $y=-s$. Now compute

$$zw=xr -ys +i(xs +yr)$$

$zw$ is real provided that $xs+yr=(x-r)y=0$.

If $y=0$ then $s=0$ and both $z$ and $w$ are real.

If $x=r$ we have $z=\bar{w}$

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I'll just explain this with barely any algebra because this just shows how beautiful complex numbers are.

First of all, when you multiply two complex numbers, you rotate by adding the angles and dialate by multiplying their moduluses: $$r_1e^{i\theta_1}\cdot r_2e^{i\theta_2}=(r_1r_2)e^{i(\theta_1+\theta_2)}$$ So, for a complex number to be real, it has to have $\theta_1+\theta_2=\pi n$ for any integer $n$, but taking the principal we'll just look at $0$ and $\pi$, so we either need $\theta_1=-\theta_2$ or $\theta_1=\pi-\theta_2$. The first equation reflects on the $x$-axis while the second reflects on the $y$-axis. (with greater or smaller magnitude)

Now when you add two complex numbers, you translate one by a vector corresponding to the other. Now, if you noticed, if you tried to translate a complex number by it's reflection on the $y$-axis, it would always be away from the $x$-axis, and with some trivial geometry you can find out why, so we need the reflection to be on the $x$-axis.

For such two complex numbers, the translation to end up on the $x$-axis and be a real number (considering the discussion of the angles above), you have to have the other complex number be of equal length, and hence the conclusion. (a conjugate complex number is a reflection on the $x$-axis with equal length)

I suggest you study closely how complex numbers work, it will help you visualize them and understand how much they're useful at the core before just wanting to do the algebra.

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  • $\begingroup$ Can anyone help me why my answer is getting downvoted? $\endgroup$ May 26, 2020 at 19:27

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