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Is it possible that a polynomial of degree $n$ with real coefficients has exactly one complex root? I saw https://en.wikipedia.org/wiki/Complex_conjugate_root_theorem but wondered if this can happen for polynomials several variables.

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  • $\begingroup$ The polynomial $z^n$... $\endgroup$ – Zev Chonoles Jul 14 '15 at 15:47
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    $\begingroup$ This is easily answered by considering the single variable case, where the answer is no. Specifically, for $n-1$ of the variables, choose any real values, and then you are solving for a single variable polynomial, which will have multiple complex roots if it has any at all. $\endgroup$ – rondo9 Jul 14 '15 at 15:49
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What do you mean with complex root? If you consider real numbers as particular complex numbers ($i.e.$ complex numbers with zero-imaginary part) then every polynomial of the form $p(z)=(z-z_0)^n$ where $z_0$ is a real number has exactly one complex root.

But if you mean a complex number which is not real; then this is impossible. Indeed, let $z_0=a+ib$ be any complex number; then the only polynomials which have $z_0$ as unique complex solution are $p(z)=(z-z_0)^n$ where $n$ is a non-zero natural number. But clearly, in this case, your polynomial has not real coefficients.

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This is technically possible if you consider the root of a real valued line is a one real number, and all real numbers are complex.

Edit: this doesn't violate your theorem since any real number is also it's complex conjugate

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