0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ If it is not $1$, we can divide it so that it becomes $1$. $\endgroup$ – Idonknow Oct 18 '17 at 14:19
  • $\begingroup$ Also, it's the fundamental theorem of algebra. $\endgroup$ – Randall Oct 18 '17 at 14:20
4
$\begingroup$

$P(z) = 0$ 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 $1$.

$\endgroup$
  • $\begingroup$ beat me to it.. $\endgroup$ – Gregory Oct 18 '17 at 14:19
  • $\begingroup$ Well I'm sitting in office hours now so I'm super bored (but I shouldn't be). $\endgroup$ – Randall Oct 18 '17 at 14:22
  • $\begingroup$ 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 $\endgroup$ – Gregory Oct 18 '17 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.