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Post by Admin on Mar 2, 2020 9:42:36 GMT
If $c$ is a root of $P_n(z)$ then
$0=P_n(c)=(c-c_1)(c-c_2)...(c-c_n)$
Since $0=P_n(c)$ then $0=\overline{P_n(c)}=(\overline{c-c_1})(\overline{c-c_2})...(\overline{c-c_n})=(\overline{c}-\overline{c_1})(\overline{c}-\overline{c_2})...(\overline{c}-\overline{c_n})$.
Since the coefficients $c_1,c_2,...,c_n$ are assumed to be real then we can write this as
$0=P_n(c)=(\overline{c}-c_1)(\overline{c}-c_2)...(\overline{c}-c_n)=P_n(\overline{c})$.
So if $c$ is a root of $P_n(z)=0$ then so is $\overline{c}$.
Vasco/Admin
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Mar 2, 2020 21:27:24 GMT
If $c$ is a root of $P_n(z)$ then $0=P_n(c)=(c-c_1)(c-c_2)...(c-c_n)$ Since $0=P_n(c)$ then $0=\overline{P_n(c)}=(\overline{c-c_1})(\overline{c-c_2})...(\overline{c-c_n})=(\overline{c}-\overline{c_1})(\overline{c}-\overline{c_2})...(\overline{c}-\overline{c_n})$. Since the coefficients $c_1,c_2,...,c_n$ are assumed to be real then we can write this as $0=P_n(c)=(\overline{c}-c_1)(\overline{c}-c_2)...(\overline{c}-c_n)=P_n(\overline{c})$. So if $c$ is a root of $P_n(z)=0$ then so is $\overline{c}$. Vasco/Admin Vasco,
But on the same page it says that $P_n(c)$ can only be counted on to have n roots if one admits complex numbers, so can one assume the coefficients to be real?
Gary
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Post by Admin on Mar 3, 2020 0:49:42 GMT
If $c$ is a root of $P_n(z)$ then $0=P_n(c)=(c-c_1)(c-c_2)...(c-c_n)$ Since $0=P_n(c)$ then $0=\overline{P_n(c)}=(\overline{c-c_1})(\overline{c-c_2})...(\overline{c-c_n})=(\overline{c}-\overline{c_1})(\overline{c}-\overline{c_2})...(\overline{c}-\overline{c_n})$. Since the coefficients $c_1,c_2,...,c_n$ are assumed to be real then we can write this as $0=P_n(c)=(\overline{c}-c_1)(\overline{c}-c_2)...(\overline{c}-c_n)=P_n(\overline{c})$. So if $c$ is a root of $P_n(z)=0$ then so is $\overline{c}$. Vasco/Admin Vasco,
But on the same page it says that $P_n(c)$ can only be counted on to have n roots if one admits complex numbers, so can one assume the coefficients to be real?
Gary
Gary Rather it says, near the top of page 24, that if we don't admit complex numbers we can't factorise a polynomial of degree $n$ into a product of $n$ factors. Then, on the same page, where the suggested exercise is posed, it says "For if the coefficients ... are all real then $P_n(z)=0$ implies [exercise] $P_n(\overline{c})=0$,... Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 3, 2020 1:42:52 GMT
Vasco,
But on the same page it says that $P_n(c)$ can only be counted on to have n roots if one admits complex numbers, so can one assume the coefficients to be real?
Gary
Gary Rather it says, near the top of page 24, that if we don't admit complex numbers we can't factorise a polynomial of degree $n$ into a product of $n$ factors. Than, on the same page, where the suggested exercise is posed, it says "For if the coefficients ... are all real then $P_n(z)=0$ implies [exercise] $P_n(\overline{c})=0$,... Vasco Vasco,
Sorry for the misdirection. I see it now.
Gary
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