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Totally new here on the site so... I'm having a bit of trouble with determining on whether or not a set of complex numbers is a line. The following is an example:

Consider the set of all complex numbers z that satisfy the equation...
$$ 5z-7\bar{z} = 4+2i $$
Show whether or not it's a line in the complex plane or not.

Completely stuck on where to even begin much less solve it/prove it.
Any help would be appreciated.

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Basic approach. Let $z = a+bi$ and then $\overline{z} = a-bi$. In that case, we have

$$ 5(a+bi)-7(a-bi) = 4+2i $$

Solve for $a$ and $b$.

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Let $z=x+iy.$ Then $\overline z=x-iy$. Now put this into your equation, you get $$5(x+iy)-7(x-iy)=4+2i$$.

So, $$-2x+i\times 12y=4+2i $$

Compare, the real and imaginary parts, you get $$-2x=4\\\implies x=-2$$ and $$12y=2\\\implies y=\frac 16.$$

So, $$z=-2+\frac i6$$

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  • $\begingroup$ Solving for unknowns I pretty well understand. It's proving whether or not this equation is a line in the complex plane I'm having a bit of trouble with. $\endgroup$ – user309502 Jan 31 '16 at 8:18
  • $\begingroup$ @Romana, this is just a point satisfying the above equation. A line contains infinitely many points, not only one point. So, this is not a line in the complex plane. $\endgroup$ – user249332 Jan 31 '16 at 8:26
  • $\begingroup$ @Subhadeep_Dey, Oh goodness, I completely forgot about that, haha. So how would I be able to show whether or not an equation (obviously not this one) is a line in the complex plane or not? Would solving be beneficial? Or am I missing something in my thinking? $\endgroup$ – user309502 Jan 31 '16 at 8:42
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    $\begingroup$ @Romana: Subhadeep_Dey's solution, and mine (although I left it more implicit), are fully general, and result in a single point—i.e., specific values for the real and imaginary components of $z$. If you could solve for $x$ and $y$ and obtain, in place of these specific values, an equation of the form $y = mx+b$, that would demonstrate that the valid values of $z$ fall on a line in the complex plane. (Of course, it may not always be that straightforward.) $\endgroup$ – Brian Tung Feb 1 '16 at 16:59

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