For each $ a $ with $ |a| > 1 $, $f^{-1}(a)$ contains exactly one point
$begingroup$
Let $ G = { z in mathbb{C} : |z-2| < 1} $ and let $ f $ be
analytic in $ overline{G} $ except for one simple pole $ z_0 $
inside. Suppose that $ |f(z)| = 1$ for all $ z in partial G $. Show
that for all $ a $ with $ |a| > 1 $, $ f^{-1}(a) $ contains exactly
one point.
So $f(z) $ can be written as $ sum a_n (z-2)^n + frac{b}{z - z_0} $, but I don't know how to use the fact that $ |f(z)| = 1 $ for all $ z in partial G $. How to solve this question?
complex-analysis
asked Jan 7 at 10:01
lifeishard911lifeishard911
1166
$endgroup$
add a comment |
$begingroup$
Let $ G = { z in mathbb{C} : |z-2| < 1} $ and let $ f $ be
analytic in $ overline{G} $ except for one simple pole $ z_0 $
inside. Suppose that $ |f(z)| = 1$ for all $ z in partial G $. Show
that for all $ a $ with $ |a| > 1 $, $ f^{-1}(a) $ contains exactly
one point.
So $f(z) $ can be written as $ sum a_n (z-2)^n + frac{b}{z - z_0} $, but I don't know how to use the fact that $ |f(z)| = 1 $ for all $ z in partial G $. How to solve this question?
complex-analysis
asked Jan 7 at 10:01
lifeishard911lifeishard911
1166
$endgroup$
add a comment |
$begingroup$
Let $ G = { z in mathbb{C} : |z-2| < 1} $ and let $ f $ be
analytic in $ overline{G} $ except for one simple pole $ z_0 $
inside. Suppose that $ |f(z)| = 1$ for all $ z in partial G $. Show
that for all $ a $ with $ |a| > 1 $, $ f^{-1}(a) $ contains exactly
one point.
So $f(z) $ can be written as $ sum a_n (z-2)^n + frac{b}{z - z_0} $, but I don't know how to use the fact that $ |f(z)| = 1 $ for all $ z in partial G $. How to solve this question?
complex-analysis
asked Jan 7 at 10:01
lifeishard911lifeishard911
1166
$endgroup$
Let $ G = { z in mathbb{C} : |z-2| < 1} $ and let $ f $ be
analytic in $ overline{G} $ except for one simple pole $ z_0 $
inside. Suppose that $ |f(z)| = 1$ for all $ z in partial G $. Show
that for all $ a $ with $ |a| > 1 $, $ f^{-1}(a) $ contains exactly
one point.
So $f(z) $ can be written as $ sum a_n (z-2)^n + frac{b}{z - z_0} $, but I don't know how to use the fact that $ |f(z)| = 1 $ for all $ z in partial G $. How to solve this question?
complex-analysis
complex-analysis
asked Jan 7 at 10:01
lifeishard911lifeishard911
1166
asked Jan 7 at 10:01
lifeishard911lifeishard911
1166
asked Jan 7 at 10:01
lifeishard911lifeishard911
1166
asked Jan 7 at 10:01
lifeishard911lifeishard911
1166
asked Jan 7 at 10:01
lifeishard911lifeishard911
1166
1166
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since this is complex analysis, $ |f(z)| = 1 $ on $ partial G $ should be interpreted in one of two different ways :
For the maximum modulus principle, however this result doesn't help at all with finding only one preimage of $ a $.
For the indices of points relative to some curves : here, we get by setting the usual curve $ gamma : t mapsto e^{2i pi t} + 2 $ that $ n(f circ gamma, a) = 0 $ because $ operatorname{Im} f circ gamma subset D_c(0, 1)$ and $ a not in D_c(0, 1)$. Thus , we can use the argument principle :
$$ int_gamma frac{f'(z)}{f(z) - a}dz = 2 i pi (sum n(gamma, text{zero of } f(z) - a) - sum n(gamma, text{pole of } f(z) - a))$$
and notice that $ int_gamma frac{f'(z)}{f(z) - a}dz = n(fcirc gamma, a) = 0$. The result then follows.
answered Jan 7 at 10:53
FreeSaladFreeSalad
454211
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Since this is complex analysis, $ |f(z)| = 1 $ on $ partial G $ should be interpreted in one of two different ways :
For the maximum modulus principle, however this result doesn't help at all with finding only one preimage of $ a $.
For the indices of points relative to some curves : here, we get by setting the usual curve $ gamma : t mapsto e^{2i pi t} + 2 $ that $ n(f circ gamma, a) = 0 $ because $ operatorname{Im} f circ gamma subset D_c(0, 1)$ and $ a not in D_c(0, 1)$. Thus , we can use the argument principle :
$$ int_gamma frac{f'(z)}{f(z) - a}dz = 2 i pi (sum n(gamma, text{zero of } f(z) - a) - sum n(gamma, text{pole of } f(z) - a))$$
and notice that $ int_gamma frac{f'(z)}{f(z) - a}dz = n(fcirc gamma, a) = 0$. The result then follows.
answered Jan 7 at 10:53
FreeSaladFreeSalad
454211
$endgroup$
add a comment |
$begingroup$
Since this is complex analysis, $ |f(z)| = 1 $ on $ partial G $ should be interpreted in one of two different ways :
For the maximum modulus principle, however this result doesn't help at all with finding only one preimage of $ a $.
For the indices of points relative to some curves : here, we get by setting the usual curve $ gamma : t mapsto e^{2i pi t} + 2 $ that $ n(f circ gamma, a) = 0 $ because $ operatorname{Im} f circ gamma subset D_c(0, 1)$ and $ a not in D_c(0, 1)$. Thus , we can use the argument principle :
$$ int_gamma frac{f'(z)}{f(z) - a}dz = 2 i pi (sum n(gamma, text{zero of } f(z) - a) - sum n(gamma, text{pole of } f(z) - a))$$
and notice that $ int_gamma frac{f'(z)}{f(z) - a}dz = n(fcirc gamma, a) = 0$. The result then follows.
answered Jan 7 at 10:53
FreeSaladFreeSalad
454211
$endgroup$
add a comment |
$begingroup$
Since this is complex analysis, $ |f(z)| = 1 $ on $ partial G $ should be interpreted in one of two different ways :
For the maximum modulus principle, however this result doesn't help at all with finding only one preimage of $ a $.
For the indices of points relative to some curves : here, we get by setting the usual curve $ gamma : t mapsto e^{2i pi t} + 2 $ that $ n(f circ gamma, a) = 0 $ because $ operatorname{Im} f circ gamma subset D_c(0, 1)$ and $ a not in D_c(0, 1)$. Thus , we can use the argument principle :
$$ int_gamma frac{f'(z)}{f(z) - a}dz = 2 i pi (sum n(gamma, text{zero of } f(z) - a) - sum n(gamma, text{pole of } f(z) - a))$$
and notice that $ int_gamma frac{f'(z)}{f(z) - a}dz = n(fcirc gamma, a) = 0$. The result then follows.
answered Jan 7 at 10:53
FreeSaladFreeSalad
454211
$endgroup$
Since this is complex analysis, $ |f(z)| = 1 $ on $ partial G $ should be interpreted in one of two different ways :
For the maximum modulus principle, however this result doesn't help at all with finding only one preimage of $ a $.
For the indices of points relative to some curves : here, we get by setting the usual curve $ gamma : t mapsto e^{2i pi t} + 2 $ that $ n(f circ gamma, a) = 0 $ because $ operatorname{Im} f circ gamma subset D_c(0, 1)$ and $ a not in D_c(0, 1)$. Thus , we can use the argument principle :
$$ int_gamma frac{f'(z)}{f(z) - a}dz = 2 i pi (sum n(gamma, text{zero of } f(z) - a) - sum n(gamma, text{pole of } f(z) - a))$$
and notice that $ int_gamma frac{f'(z)}{f(z) - a}dz = n(fcirc gamma, a) = 0$. The result then follows.
answered Jan 7 at 10:53
FreeSaladFreeSalad
454211
edited Jan 7 at 13:51
answered Jan 7 at 10:53
FreeSaladFreeSalad
454211
answered Jan 7 at 10:53
FreeSaladFreeSalad
454211
answered Jan 7 at 10:53
FreeSaladFreeSalad
454211
454211
add a comment |
add a comment |
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