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Solve this equation: $$z+ \bar{z}=|z^2+1|$$

I tried the following. $$x+iy+x-iy=|z^2+1|$$ $$2x=|z^2+1|$$ $$x=(|z^2+1|)/2$$ and I came to a dead end.

How can I proceed?

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    $\begingroup$ $|z^2+1| \neq |x^2+1|$ $\endgroup$ Commented Oct 12, 2013 at 13:50
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    $\begingroup$ Breaking up $z$ into its real and imaginary parts is a good approach, you just need to continue in that fashion on the right hand side, e.g. $|z^2 + 1| = \ldots$ ? $\endgroup$
    – hardmath
    Commented Oct 12, 2013 at 13:53
  • $\begingroup$ sry is yet another typo. $\endgroup$
    – malloc
    Commented Oct 12, 2013 at 13:55

2 Answers 2

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Instead of splitting into real and imaginary parts, one can also square the original equation and get $$ (z+\bar z)^2 = (z^2+1)\overline{(z^2+1)} = (z^2+1)(\bar z^2+1) $$ (where the second equals sign is because conjugation is an isomorphism).

After multiplying out, the $z^2$ and $\bar z^2$ terms cancel out and we're left with $$ (z\bar z)^2 - 2z\bar z + 1 = 0 $$ which factors as $$ (z\bar z-1)^2 = 0 $$


Here's a diagram that shows geometrically that the equation is indeed true on the right half of the unit circle. The three congruent right triangles show that the distance to $z^2+1$ is twice the real part of $z$.

diagram goes here

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Hints: $$z^2+1=(x+iy)^2+1=(x^2-y^2+1)+i(2xy)$$ $$|a+ib|^2=a^2+b^2$$

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    $\begingroup$ @malloc: I'm afraid it isn't that simple, no. Squaring both sides of $$2x=|z^2+1|$$ will give you $$4x^2=(x^2-y^2+1)^2+4x^2y^2,$$ which is more complicated. $\endgroup$ Commented Oct 12, 2013 at 14:09
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    $\begingroup$ @malloc: Henning makes a good point. Once you expand $(x^2-y^2+1)^2$ and gather all your variables on one side, you should be able to show that the expression on one side is a perfect square, while the expression on the other is $0.$ Taking the square root will give you your solution: a circle. $\endgroup$ Commented Oct 12, 2013 at 14:22
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    $\begingroup$ @malloc: Perfect! So you have $(x^2+y^2-1)^2=0,$ so $x^2+y^2-1=0,$ and so $x^2+y^2=1.$ Since $x$ is non-negative by the original equation, what can you conclude? $\endgroup$ Commented Oct 12, 2013 at 14:38
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    $\begingroup$ @malloc: The plot on Wolfram is under the assumption that $z$ is real. The only solution there is $z=1.$ Ignore it. $\endgroup$ Commented Oct 12, 2013 at 14:47
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    $\begingroup$ @malloc: By the way, even if you don't accept Henning's answer (though I would, if I were you), I definitely recommend that you take a good look at it. That approach is much less complicated than splitting into parts, and makes the process almost trivial. $\endgroup$ Commented Oct 12, 2013 at 14:51

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