Exercise problem. I do not need a full solution because I am trying to solve myself. Just a hint would be great. Let $f$ be a polynomial with real coefficients. How to show that $f(z) = f(i)$ for every complex $z$?
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Checked it. I typed in the problem statement word by word exactly like in the book. – fade2black Jun 27 '13 at 20:41
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@fade2black What's the book? – Git Gud Jun 27 '13 at 20:41
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Was it perhaps "with real values" instead of "with real coefficients"? – Daniel Fischer Jun 27 '13 at 20:43
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@DanielFischer Nice guess. – Git Gud Jun 27 '13 at 20:44
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Tom M. Apostol, Calculus (Volume 1), 2nd Edition. – fade2black Jun 27 '13 at 20:45
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@fade2black What section is it in? – Git Gud Jun 27 '13 at 20:47
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@Git Gud 9.Complex Numbers – fade2black Jun 27 '13 at 20:49
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@fade2black I found it. It's wrong as Panda pointed out. Reading $(b)$ I guess the problem is supposed to be to show that $f(z)=f(\overline z)$. – Git Gud Jun 27 '13 at 20:50
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@GitGud A typo in the book, if I understand correctly? – Daniel Fischer Jun 27 '13 at 20:52
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@DanielFischer Yeah. I edited my comment above. Read it. – Git Gud Jun 27 '13 at 20:52
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3@GitGud That should be $f(z) = \overline{f(\overline{z})}$. – Daniel Fischer Jun 27 '13 at 20:53
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@Panda Turn your comment into an answer so this doesn't show as unanswered? – Git Gud Jun 27 '13 at 20:53
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@Daniel Both our RHSs are equal. – Git Gud Jun 27 '13 at 20:54
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@Daniel Fische Do you mean by "with real values" that the polynomial takes only real values? – fade2black Jun 27 '13 at 20:54
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1@GitGud I see only one conjugation in yours, and two conjugations in mine. – Daniel Fischer Jun 27 '13 at 20:55
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@fade2black Yes, that's what I meant. That would imply that it is constant. – Daniel Fischer Jun 27 '13 at 20:55
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@DanielFischer Oops. – Git Gud Jun 27 '13 at 20:57
1 Answers
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I didn't understand quite well the answer for a) in the book. But my approach (for the part a ) is to use the properties $\overline{z_1 + z_2} = \overline z_1 + \overline z_2 $ and $\overline{z_1*z_2} = \overline{z_1}*\overline{z_2}$, and induction on degree of polynomials.
Base step: $deg(f) = 0$. Let $f(x) = a$. Then $f(z) = a = \overline a = \overline{f(\overline{z})}.$
Induction: $f(x) = a_nx^n + \hat{f}(x)$, where $deg(\hat{f}) < n$.
$f(z) = a_nz^n + \hat{f}(z)$
$\overline{f(\overline{z})} = \overline{a_n(\overline z)^n + \hat{f}(\overline{z})} = \overline{a_n(\overline z)^n} + \overline{\hat{f}(\overline{z})} = a_n\overline{(\overline{z})^n} + \overline{\hat{f}(\overline{z})} = a_n(\overline{\overline{z}})^n + \overline{\hat{f}(\overline{z})} = a_nz^n + \overline{\hat{f}(\overline{z})}$.
But $\hat{f}(z)=\overline{\hat{f}(\overline{z})}$ by assumption. So, $f(z) = \overline{f(\overline{z})}$.
fade2black
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