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I was reading this proof and I was having trouble understanding the last step. The assumption is we have a entire function $ f $ such that $ |f| = 1 $ on $ |z| = 1 $. Then, I followed to proof to derive $ f(z) = C \prod\limits_{j=1}^{n} \frac{z-a_j}{1-\bar{a}z} $. Where $a_j $ are the zeroes of $ f $. The author then states that:

So we can write $f(z)=c\prod_{j=1}^n\frac{z-a_j}{1-\bar a_jz}$, and we are reduced to show that $a_j=0$ for all $j$. As $f$ is assumed entire, we need to remove the singularities in $\frac 1{\bar{a_j}}$, which has modulus $>1$. These one can't be canceled unless $a_j=0$.

I just don't understand how he concluded $ a_j = 0 $

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Max
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If $|a_j|<1,a_j\neq 0$, let $w=\frac{z-a_j}{1-\bar{a}_jz}$. Then $z=\frac{a_j+w}{1+\bar{a}_jw}$ and hence $f(z)$ becomes a new function $F(w)=f(\frac{w+a_j}{1+\bar{a}_jw})$ which is still analytic in $|w|\le1$ and $F(0)=f(a_j)=0$.

xpaul
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