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I need to solve the complex equation $z^3\bar{z}+4i=2z(z+i\bar{z})$

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z^3z?? is it equal to z^4? –  quanta May 17 '11 at 18:45
Did you mean $z^3\bar{z}+4i=2z(z+i\bar{z})$? –  Ross Millikan May 17 '11 at 18:46
@Katy23: If you meant $z^3z$, then why did you write it as $z^3z$ and not $z^4$? –  Arturo Magidin May 17 '11 at 19:04
@Katy23, where did you exactly read that? Some of us are finding it hard to believe that anyone would exactly write that. –  Gerry Myerson May 18 '11 at 0:04
If this is homework, please add the homework tag and read meta.math.stackexchange.com/questions/1803/… for guidelines on how to ask a homework question (in particular, show your work so far). If not, please describe your motivation for considering this equation, and anything you have learned about it. Others: Please consider, in the larger sense, how your answers will help the asker. –  Nate Eldredge May 18 '11 at 3:41

3 Answers 3

Solution to Ross version of the question:

$$z^3 \bar{z} + 4 i = 2z(z + i \bar{z})$$ $$z^3 \bar{z} + 4 i - 2z^2 - 2 i z\bar{z} = 0$$ $$z^2 \left(z \bar{z} - 2 \right) -2 i \left( z \bar{z} - 2 \right) = 0$$ $$\left(z^2 -2 i \right) \left(z \bar{z} - 2 \right) = 0$$ This gives us $\left(z^2 -2 i \right) = 0$ or $\left(z \bar{z} -2 \right) = 0$

Note that $\left(z^2 -2 i \right) = 0$ also satisfies $\left(z \bar{z} -2 \right) = 0$

Hence, the solution is a circle centered at the origin in the complex plane with radius $\sqrt{2}$

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The solutions to Ross's version of the equation form the circle $|z|^2 = 2$.

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I think if he just wanted the solution he would have used wolframalpha –  Listing May 17 '11 at 19:02
How to solve this equation? –  Katy23 May 17 '11 at 19:54
In this version, it becomes $z^2|z|^2+4i=2z^2+2i|z|^2$, so $|z|^2=\frac{2z^2-4i}{z^2-2i}=2$ –  Ross Millikan May 17 '11 at 21:54

For the original equation, Alpha gives $\pm \sqrt{2}, \pm(.542931+.7763264i)$

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Thanks but i need all steps to solution –  Katy23 May 17 '11 at 20:30
Looks like Alpha is doing it numerically. It doesn't offer a derivation. Because the equation only uses $z^2$, you can define $z^2=a+bi, a,b$ real, and get two quadratic equations in two unknowns. This will give a quartic, but those are hard. –  Ross Millikan May 17 '11 at 20:38
@Ross: I think you missed parenthesis on the RHS in the Alpha computation: "solve z^4+4i==2(z^2+iabs(z^2)) for z" instead of "solve z^4+4i==2z^2+iabs(z^2) for z". The computed solutions become $z=\pm\sqrt{2}$ and $z=\pm(0.542931+0.776326\; i)$. But this second pair does not satisfy $|z|=2$, as derived in your comment to Robert Israel's answer. I have no explanation. –  Américo Tavares May 17 '11 at 23:43
@Americo: you are correct, I had lost a factor 2 on the last term on the right. I have corrected. I don't think the roots should necessarily satisfy $|z|^2=2$, as there I was solving with the last $z$ on the left conjugated. –  Ross Millikan May 18 '11 at 0:13
@Ross: So one would have to solve the "two quadratic equations in two unknowns". –  Américo Tavares May 18 '11 at 0:28

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