Let $r\in\mathbb{R}$, $p\in\mathbb{C}\left[ x\right] $ and $q\left( x\right) =p\left( ix\right) $. The following reasoning is false: $p\left( r\right) =0$ iff $p\left( i\left( -ir\right) \right) =0$ iff $q\left( -ir\right) =0$; but $q\in\mathbb{C}\left[ x\right] $ so $q\left( ir\right) =0$. This is contradictory since $q\left( ir\right) =p\left( -r\right) $ however $-r\in\mathbb{R}$ is not necessarily a root of $p$. Where's the blunder?
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Why if $q(-ir)=0$ and $q\in\mathbb{C}[x]$, you deduce $q(ir)=0$? For example $q(x):=x+ir\in\mathbb{C}[x]$, for $r\neq0$, satisfies $q(-ir)=0$, but $q(ir)=2ir\neq0$. – OR. Jan 21 '14 at 12:17
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However complex roots appear in conjugated pairs. – Juan Ignacio Jan 22 '14 at 09:59
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You are remembering only half of the result. The statement should be: For polynomials with real coefficients roots appear in conjugate pairs. – OR. Jan 22 '14 at 12:27