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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1$|z^2+1| \neq |x^2+1|$ – Prahlad Vaidyanathan Oct 12 '13 at 13:50
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1Breaking 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. $|z^2 + 1| = \ldots$ ? – hardmath Oct 12 '13 at 13:53
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sry is yet another typo. – malloc Oct 12 '13 at 13:55
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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 $z$.
hmakholm left over Monica
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Hints: $$z^2+1=(x+iy)^2+1=(x^2-y^2+1)+i(2xy)$$ $$|a+ib|^2=a^2+b^2$$
Cameron Buie
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if I understand your suggestion then the result should be $ x ^ 2 + y ^ 2 +2 x +1 = $ 0 or a circle, right? – malloc Oct 12 '13 at 14:04
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@malloc: How do you get that? I end up with a fourth-degree curve. – hmakholm left over Monica Oct 12 '13 at 14:08
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1@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. – Cameron Buie Oct 12 '13 at 14:09
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No. $(2xy)^2=4x^2y^2$. The $i$ is not included in the squaring. If it were as you say, then (for example) $$1=1^2=|i|^2=i^2=-1.$$ – Cameron Buie Oct 12 '13 at 14:16
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1@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. – Cameron Buie Oct 12 '13 at 14:22
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Actually, I take everything back. I had an algebra mistake along the way. – hmakholm left over Monica Oct 12 '13 at 14:34
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I'm still a little confused(sry I'm a newbie :)). I did all the calculations and finally get $$(x^2+y^2-1)^2$$ that should be correct! Now what I'm missing? – malloc Oct 12 '13 at 14:36
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1@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? – Cameron Buie Oct 12 '13 at 14:38
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ahhhhhh ok got it! but the circle on walfram is not too big? It should have $r=1$! – malloc Oct 12 '13 at 14:42
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1@malloc: The plot on Wolfram is under the assumption that $z$ is real. The only solution there is $z=1.$ Ignore it. – Cameron Buie Oct 12 '13 at 14:47
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2@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. – Cameron Buie Oct 12 '13 at 14:51
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