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The question asks me to show that "in general," $$z \overline{z} + az + \overline{b}z + c = 0$$ has two solutions, and asks when it has more than two.

Its obvious that if $a + b = 0$ and $c$ is negative, we get an equation of a circle.

Now, the two solutions i assume come from the fact that the equation looks very similar to a quadratic polynomial in $z$. I am just having trouble connecting $\|z\|^{2} + az + \overline{b}z + c = 0$ to $z^{2} + az + \overline{b}z +c = 0$

Thanks.

Gerry Myerson
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Geano
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    Hint: if $$z \overline{z} + az + \overline{b}z + c = 0$$ then $$\overline{z \overline{z} + az + \overline{b}z + c} = 0$$ and we get $$z \overline{z} + \overline{a}\overline{z} + b\overline{z} + \overline{c} = 0$$ – Ivo Terek Oct 05 '14 at 01:36
  • Was your equation supposed to be $z\bar{z} + a z + b \bar{z} + c = 0$? –  Oct 05 '14 at 02:16

1 Answers1

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The issue of $a$ and $\overline{b}$ is not important; you really just have $|z|^2 + a z + c = 0$. Letting $a=a_1+ia_2,c=c_1+ic_2$, we have

$$x^2 + y^2 + a_1 x - a_2 y + c_1 + i (a_1 y + a_2 x + c_2) = 0$$

or

$$x^2 + y^2 + a_1 x - a_2 y + c_1 = 0, \\ a_1 y + a_2 x + c_2 = 0.$$

Complete the square in the first equation:

$$(x+a_1/2)^2 + (y-a_2/2)^2 + (c_1 - a_1^2/4 - a_2^2/4) = 0$$

So now our first equation has solution either a circle, one point, or no solutions. The second equation has solution of a line or the whole plane. The intersections are either a circle, two points, one point, or none. The "generic cases" (where we assume inequalities but no equalities) are two points or no points.

Ian
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  • clear and simple, thanks. Is turning a complex eq. into two real valued ones a common thing? – Geano Oct 05 '14 at 02:19
  • @Geano Typically if you can exploit the fact that the function is holomorphic, you do not want to do this. But here the function is not holomorphic, so there's really very little to be gained by keeping it in complex form. – Ian Oct 05 '14 at 02:45