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This is related to the question I asked regarding finding the complex roots of $z^3+\bar{z}=0$. It turned out that there were 5 complex roots, but because the equation was of degree 3 I was only expecting 3 roots at most.

Could someone explain in detail whatever it is I'm not understanding here?

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The equation you azked about involved the conjugate of $z$, so it easn't a polynomial equation. –  hardmath May 16 '12 at 6:36
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As Gerry Myerson, pointed out you should not really be surprised that $z^3 + \bar{z} = 0$ has more than $3$ roots since it is not a polynomial in $z$. For instance, the simpler equation $z+\bar{z} = 0$ has infinite roots. –  user17762 May 16 '12 at 6:37
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up vote 6 down vote accepted

As observed already, you don't have there an equation of the form $P(z)=0$ for some polynomial $P$, but rather an equation of the form $$ P(z,\bar z)=0\qquad\qquad(\star) $$ where $P$ is a polynomial in two variables of degree, say, $d$. After the substitutions $z=x+iy$ and $\bar z=x-iy$, and separating the real from the imaginary part, equation $(\star)$ is transformed into a system $$ P_1(x,y)=P_2(x,y)=0\qquad\qquad(\star\star) $$ where $P_1$ and $P_2$ are polynomial with real coefficients and you are looking for the real solutions of $(\star\star)$.

Since $\deg(P_1)=\deg(P_2)=\deg(P)=d$, if $(\star\star)$ does not admit infinitely many solutions (which may as well happen, see the comment of Marvis) you may expect up to $d^2$ solutions by Bezout's Theorem.

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The concept of degree applies only to polynomials, and the presence of the complex conjugate means you're not talking about polynomials, so you can't talk about the degree of that equation, any more than you can talk about the degree of the equation $\sin x=0$.

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