Let $f(x)$ and $g(x)$ are two complex polynomials such that $f^{-1}(c_{i})=g^{-1}(c_{i})$
$begingroup$
Let $f(x)$ and $g(x)$ are two complex polynomials such that $f^{-1}(c_{i})=g^{-1}(c_{i})$ for two distinct complex numbers $c_{i}$, $i=1,2$.
Then can we say $f=g$?
Here nothing is given about the multiplicity $f^{-1}(c_{i})$. So any hint please..
Thank you.
complex-analysis polynomials
edited Dec 9 '18 at 11:28
Martin R
28.8k33356
asked Dec 8 '18 at 17:32
JOHNJOHN
485
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ and $g(x)$ are two complex polynomials such that $f^{-1}(c_{i})=g^{-1}(c_{i})$ for two distinct complex numbers $c_{i}$, $i=1,2$.
Then can we say $f=g$?
Here nothing is given about the multiplicity $f^{-1}(c_{i})$. So any hint please..
Thank you.
complex-analysis polynomials
edited Dec 9 '18 at 11:28
Martin R
28.8k33356
asked Dec 8 '18 at 17:32
JOHNJOHN
485
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ and $g(x)$ are two complex polynomials such that $f^{-1}(c_{i})=g^{-1}(c_{i})$ for two distinct complex numbers $c_{i}$, $i=1,2$.
Then can we say $f=g$?
Here nothing is given about the multiplicity $f^{-1}(c_{i})$. So any hint please..
Thank you.
complex-analysis polynomials
edited Dec 9 '18 at 11:28
Martin R
28.8k33356
asked Dec 8 '18 at 17:32
JOHNJOHN
485
$endgroup$
Let $f(x)$ and $g(x)$ are two complex polynomials such that $f^{-1}(c_{i})=g^{-1}(c_{i})$ for two distinct complex numbers $c_{i}$, $i=1,2$.
Then can we say $f=g$?
Here nothing is given about the multiplicity $f^{-1}(c_{i})$. So any hint please..
Thank you.
complex-analysis polynomials
complex-analysis polynomials
edited Dec 9 '18 at 11:28
Martin R
28.8k33356
asked Dec 8 '18 at 17:32
JOHNJOHN
485
edited Dec 9 '18 at 11:28
Martin R
28.8k33356
asked Dec 8 '18 at 17:32
JOHNJOHN
485
edited Dec 9 '18 at 11:28
Martin R
28.8k33356
edited Dec 9 '18 at 11:28
Martin R
28.8k33356
edited Dec 9 '18 at 11:28
Martin R
28.8k33356
28.8k33356
asked Dec 8 '18 at 17:32
JOHNJOHN
485
asked Dec 8 '18 at 17:32
JOHNJOHN
485
asked Dec 8 '18 at 17:32
JOHNJOHN
485
485
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Either $f$ and $g$ are both constant, or $f=g$. For a proof, assume that
(without loss of generality), $deg f ge deg g$, and consider the function
$$
h(z) = frac{f'(z)(f(z) - g(z))}{(f(z)-c_1)(f(z) - c_2)} , .
$$
Show that
$h$ has only removable singularities, and therefore can be extended
to an entire function.
$h$ is bounded, and therefore constant.
$h$ is identically zero.
Or, without using complex analysis: Every zero of $(f-c_1)(f-c_2)$ is a zero of $f'(f-g)$ with at least the same multiplicity, therefore
$$
f'(f-g) = h(f-c_1)(f-c_2)
$$
for some polynomial $h$. Now compare the degrees to conclude that $h = 0$.
This does not hold for entire functions in general. As an example,
$f(z) = e^z$ and $g(z) = e^{-z}$ have the same preimage for $c_1 = 0$, $c_2 = 1$, and $c_3 = -1$.
Rolf Nevanlinna showed 1929 that two (in $Bbb C$) meromorphic functions are identical if they have the same preimages for five distinct values, that is the so-called “Five-Value Theorem.”
answered Dec 8 '18 at 17:51
Martin RMartin R
28.8k33356
$endgroup$
$begingroup$
Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
$endgroup$
– JOHN
Dec 8 '18 at 18:46
1
$begingroup$
@JOHN: No – see update.
$endgroup$
– Martin R
Dec 8 '18 at 19:14
add a comment |
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$begingroup$
Either $f$ and $g$ are both constant, or $f=g$. For a proof, assume that
(without loss of generality), $deg f ge deg g$, and consider the function
$$
h(z) = frac{f'(z)(f(z) - g(z))}{(f(z)-c_1)(f(z) - c_2)} , .
$$
Show that
$h$ has only removable singularities, and therefore can be extended
to an entire function.
$h$ is bounded, and therefore constant.
$h$ is identically zero.
Or, without using complex analysis: Every zero of $(f-c_1)(f-c_2)$ is a zero of $f'(f-g)$ with at least the same multiplicity, therefore
$$
f'(f-g) = h(f-c_1)(f-c_2)
$$
for some polynomial $h$. Now compare the degrees to conclude that $h = 0$.
This does not hold for entire functions in general. As an example,
$f(z) = e^z$ and $g(z) = e^{-z}$ have the same preimage for $c_1 = 0$, $c_2 = 1$, and $c_3 = -1$.
Rolf Nevanlinna showed 1929 that two (in $Bbb C$) meromorphic functions are identical if they have the same preimages for five distinct values, that is the so-called “Five-Value Theorem.”
answered Dec 8 '18 at 17:51
Martin RMartin R
28.8k33356
$endgroup$
$begingroup$
Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
$endgroup$
– JOHN
Dec 8 '18 at 18:46
1
$begingroup$
@JOHN: No – see update.
$endgroup$
– Martin R
Dec 8 '18 at 19:14
add a comment |
$begingroup$
Either $f$ and $g$ are both constant, or $f=g$. For a proof, assume that
(without loss of generality), $deg f ge deg g$, and consider the function
$$
h(z) = frac{f'(z)(f(z) - g(z))}{(f(z)-c_1)(f(z) - c_2)} , .
$$
Show that
$h$ has only removable singularities, and therefore can be extended
to an entire function.
$h$ is bounded, and therefore constant.
$h$ is identically zero.
Or, without using complex analysis: Every zero of $(f-c_1)(f-c_2)$ is a zero of $f'(f-g)$ with at least the same multiplicity, therefore
$$
f'(f-g) = h(f-c_1)(f-c_2)
$$
for some polynomial $h$. Now compare the degrees to conclude that $h = 0$.
This does not hold for entire functions in general. As an example,
$f(z) = e^z$ and $g(z) = e^{-z}$ have the same preimage for $c_1 = 0$, $c_2 = 1$, and $c_3 = -1$.
Rolf Nevanlinna showed 1929 that two (in $Bbb C$) meromorphic functions are identical if they have the same preimages for five distinct values, that is the so-called “Five-Value Theorem.”
answered Dec 8 '18 at 17:51
Martin RMartin R
28.8k33356
$endgroup$
$begingroup$
Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
$endgroup$
– JOHN
Dec 8 '18 at 18:46
1
$begingroup$
@JOHN: No – see update.
$endgroup$
– Martin R
Dec 8 '18 at 19:14
add a comment |
$begingroup$
Either $f$ and $g$ are both constant, or $f=g$. For a proof, assume that
(without loss of generality), $deg f ge deg g$, and consider the function
$$
h(z) = frac{f'(z)(f(z) - g(z))}{(f(z)-c_1)(f(z) - c_2)} , .
$$
Show that
$h$ has only removable singularities, and therefore can be extended
to an entire function.
$h$ is bounded, and therefore constant.
$h$ is identically zero.
Or, without using complex analysis: Every zero of $(f-c_1)(f-c_2)$ is a zero of $f'(f-g)$ with at least the same multiplicity, therefore
$$
f'(f-g) = h(f-c_1)(f-c_2)
$$
for some polynomial $h$. Now compare the degrees to conclude that $h = 0$.
This does not hold for entire functions in general. As an example,
$f(z) = e^z$ and $g(z) = e^{-z}$ have the same preimage for $c_1 = 0$, $c_2 = 1$, and $c_3 = -1$.
Rolf Nevanlinna showed 1929 that two (in $Bbb C$) meromorphic functions are identical if they have the same preimages for five distinct values, that is the so-called “Five-Value Theorem.”
answered Dec 8 '18 at 17:51
Martin RMartin R
28.8k33356
$endgroup$
Either $f$ and $g$ are both constant, or $f=g$. For a proof, assume that
(without loss of generality), $deg f ge deg g$, and consider the function
$$
h(z) = frac{f'(z)(f(z) - g(z))}{(f(z)-c_1)(f(z) - c_2)} , .
$$
Show that
$h$ has only removable singularities, and therefore can be extended
to an entire function.
$h$ is bounded, and therefore constant.
$h$ is identically zero.
Or, without using complex analysis: Every zero of $(f-c_1)(f-c_2)$ is a zero of $f'(f-g)$ with at least the same multiplicity, therefore
$$
f'(f-g) = h(f-c_1)(f-c_2)
$$
for some polynomial $h$. Now compare the degrees to conclude that $h = 0$.
This does not hold for entire functions in general. As an example,
$f(z) = e^z$ and $g(z) = e^{-z}$ have the same preimage for $c_1 = 0$, $c_2 = 1$, and $c_3 = -1$.
Rolf Nevanlinna showed 1929 that two (in $Bbb C$) meromorphic functions are identical if they have the same preimages for five distinct values, that is the so-called “Five-Value Theorem.”
answered Dec 8 '18 at 17:51
Martin RMartin R
28.8k33356
edited Dec 10 '18 at 8:59
answered Dec 8 '18 at 17:51
Martin RMartin R
28.8k33356
answered Dec 8 '18 at 17:51
Martin RMartin R
28.8k33356
answered Dec 8 '18 at 17:51
Martin RMartin R
28.8k33356
28.8k33356
$begingroup$
Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
$endgroup$
– JOHN
Dec 8 '18 at 18:46
1
$begingroup$
@JOHN: No – see update.
$endgroup$
– Martin R
Dec 8 '18 at 19:14
add a comment |
$begingroup$
Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
$endgroup$
– JOHN
Dec 8 '18 at 18:46
1
$begingroup$
@JOHN: No – see update.
$endgroup$
– Martin R
Dec 8 '18 at 19:14
$begingroup$
Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
$endgroup$
– JOHN
Dec 8 '18 at 18:46
$begingroup$
Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
$endgroup$
– JOHN
Dec 8 '18 at 18:46
1
1
$begingroup$
@JOHN: No – see update.
$endgroup$
– Martin R
Dec 8 '18 at 19:14
$begingroup$
@JOHN: No – see update.
$endgroup$
– Martin R
Dec 8 '18 at 19:14
add a comment |
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