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Let polynomial $p(z)=z^2+az+b$ be such that $a$and $b$ are complex numbers and $|p(z)|=1$ whenever $|z|=1$. Prove that $a=0$ and $b=0$.

I could not make much progress. I let $z=e^{i\theta}$ and $a=a_1+ib_1$ and $b=a_2+ib_2$

Using these values in $P(z)$ i got $|P (z)|^2=1=(\cos (2\theta)-a_2\sin (\theta)+a_1\cos (\theta)+b_1)^2+(\sin(2\theta)+a_1\sin (\theta)+a_2\cos (\theta)+b_2)^2$

But i dont see how to proceed further neither can i think of any other approach any other approach. So, someone please help. I dont know complex analysis so it would be more helpful if someone can provide hints/solutions that dont use complex analysis.

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2 Answers 2

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For every $z\in \mathbb{C}$ we have $$ |p(z)|^2=(z^2+az+b)(\bar{z}^2+\bar{a}\bar{z}+\bar{b})=|z|^4+a\bar{z}|z|^2+\bar{a}z|z|^2+|a|^2|z|^2+\bar{b}z^2+b\bar{z}^2+a\bar{b}z+\bar{a}b\bar{z}+|b|^2, $$ in particular, when $|z|=1$, we have: $$ |p(z)|^2=\bar{b}z^2+b\bar{z}^2+(a+\bar{a}b)\bar{z}+(\bar{a}+a\bar{b})z+|a|^2+|b|^2+1. $$ Since $|p(z)|=1$ when $|z|=1$, we have: $$\tag{1} \bar{b}z^2+b\bar{z}^2+(a+\bar{a}b)\bar{z}+(\bar{a}+a\bar{b})z+|a|^2+|b|^2=0. $$ Putting $z=-1,1,-i,i$ in (1), we get: $$ \left\{ \begin{array}{lcc} \bar{b}+b-(a+\bar{a}b)-(\bar{a}+a\bar{b})+|a|^2+|b|^2&=&0\\ \bar{b}+b+(a+\bar{a}b)+(\bar{a}+a\bar{b})+|a|^2+|b|^2&=&0\\ -\bar{b}-b-i(a+\bar{a}b)-i(\bar{a}+a\bar{b})+|a|^2+|b|^2&=&0\\ -\bar{b}-b+i(a+\bar{a}b)+i(\bar{a}+a\bar{b})+|a|^2+|b|^2&=&0 \end{array}\right., $$ and combining these four identities we have: $$ 4(|a|^2+|b|^2)=0. $$ Thus $a=b=0$.

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  • $\begingroup$ Could you elaborate on how the signs on $b, \overline{b}$ differ between the two equations? I get $(-i)^2 = i^2 = -1$, so they should have the same signs, but it is late at night for me. . . $\endgroup$
    – user88319
    Commented Jul 9, 2015 at 6:12
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Hint: What happens when $ z = 1 $, $ z = i $, $z = -1$, $z = -i$ ?

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