what is the condition of a to get
$$ax_1\bar x_2+\bar a \bar x_1x_20$$ where $a,x_1,x_2 \in C$ $x_1\neq 0$ $x_2\neq 0$
Please, help

Question

what is the condition of a to get 
$$ax_1\bar x_2+\bar a \bar x_1x_2>0$$ where $a,x_1,x_2 \in C$ $x_1\neq 0$  $x_2\neq 0$
Please, help

loser66 · Answer

@phi

phi · Answer

I looked at this, and did not make much progress.  If you ever find out, please post the answer here and tag me.

loser66 · Answer

:( and :)

loser66 · Answer

@satellite73

p0sitr0n · Answer

x bar is the mean?

loser66 · Answer

conjugate of x

anonymous · Answer

Even if both $x$ values are non-zero it is still possible that they multiply to 0, isn't it?

loser66 · Answer

not sure.

anonymous · Answer

What about $(1-i)(-1+i)$

anonymous · Answer

Oh wait, I guess that one doesn't matter.

anonymous · Answer

Let me think... $$
(a+bi)(c+di) = (ac-bd) + (ad+bc)i
$$So we have: $$
ac=bd\
ad=-bc
$$This means $a = bd/c$.  so $$
bd^2/c =bc \implies bd^2=-bc^2\implies d^2=-c^2
$$The square of real numbers are positive so it really is impossible.

anonymous · Answer

I have a question....

anonymous · Answer

How can imaginary number be greater than another number?

anonymous · Answer

Is $i  > 0$?

loser66 · Answer

i cannot >0

anonymous · Answer

So this product has to be a real number...

anonymous · Answer

actually!!  That makes sense.  You are adding two conjugates.  They must add up to real number.

anonymous · Answer

Maybe we can simplify the problem to: $$
ax+\bar{a}\bar{x}>0
$$

anonymous · Answer

It must be the case that $2\Re(ax) > 0 \implies \Re(ax) > 0 $

anonymous · Answer

This means $$
\Re(a)\Im(x)-\Re(x)\Im(a) > 0
$$

loser66 · Answer

we have 3 numbers, not 2.

anonymous · Answer

I am saying let $x=x_1\bar{x_2}$

loser66 · Answer

Can you give me a particular example to make it clear? Please

anonymous · Answer

Hmmmm, um.

anonymous · Answer

I don't think it can be simplified.