# Why can one assume - without loss of generality - that $c_n = 1$ when proving the fundamental theorem of algebra?

So I'm looking at Rudin's principles of mathematical analysis theorem 8.8 (fundamental theorem of algebra):

Suppose $c_1, ..., c_n$ are complex numbers, $n\geq1,\;a_n\neq0$, $$P(z) = \sum_0^na_kz^k.$$ Then $P(z) = 0$ for some complex $z$.

The proof starts with the sentence:

Without loss of generality, assume $a_n =1$.

And I'm not entirely sure why this is even allowed. It would be great if someone could explain this to me?

• If it is not $1$, we can divide it so that it becomes $1$. – Idonknow Oct 18 '17 at 14:19
• Also, it's the fundamental theorem of algebra. – Randall Oct 18 '17 at 14:20

$P(z) = 0$ if and only if $\frac{1}{a_n}P(z)=0$ so it doesn't matter. The result of this converts the coefficient of $z^n$ to $1$.

• beat me to it.. – Gregory Oct 18 '17 at 14:19
• Well I'm sitting in office hours now so I'm super bored (but I shouldn't be). – Randall Oct 18 '17 at 14:22
• Hah. When I was a TA in grad school I preferred not having anyone showing up to complain about a point lost here or there :p – Gregory Oct 18 '17 at 14:54