Using the estimation lemma (ML inequality) to prove the following inequality.
$begingroup$
show that |∮1/z dz| ≤ (3/4) π where γ :[0, 3/8] and γ(t) = e^(−i2πt)?
I know that the ML inequality shows that:
∮f(z)dz ≤ ML(γ) but am struggling to calculate ML
complex-analysis inequality
asked Dec 13 '18 at 11:12
Aoife CoyleAoife Coyle
125
$endgroup$
add a comment |
$begingroup$
show that |∮1/z dz| ≤ (3/4) π where γ :[0, 3/8] and γ(t) = e^(−i2πt)?
I know that the ML inequality shows that:
∮f(z)dz ≤ ML(γ) but am struggling to calculate ML
complex-analysis inequality
asked Dec 13 '18 at 11:12
Aoife CoyleAoife Coyle
125
$endgroup$
add a comment |
$begingroup$
show that |∮1/z dz| ≤ (3/4) π where γ :[0, 3/8] and γ(t) = e^(−i2πt)?
I know that the ML inequality shows that:
∮f(z)dz ≤ ML(γ) but am struggling to calculate ML
complex-analysis inequality
asked Dec 13 '18 at 11:12
Aoife CoyleAoife Coyle
125
$endgroup$
show that |∮1/z dz| ≤ (3/4) π where γ :[0, 3/8] and γ(t) = e^(−i2πt)?
I know that the ML inequality shows that:
∮f(z)dz ≤ ML(γ) but am struggling to calculate ML
complex-analysis inequality
complex-analysis inequality
asked Dec 13 '18 at 11:12
Aoife CoyleAoife Coyle
125
asked Dec 13 '18 at 11:12
Aoife CoyleAoife Coyle
125
asked Dec 13 '18 at 11:12
Aoife CoyleAoife Coyle
125
asked Dec 13 '18 at 11:12
Aoife CoyleAoife Coyle
125
asked Dec 13 '18 at 11:12
Aoife CoyleAoife Coyle
125
125
add a comment |
add a comment |
1 Answer
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$begingroup$
$gamma$ is the part of the unit circle from $1$ to $-i$ and $|z|=1$ on the circle. So $M=1$. To find $L$ use the formula $L=int_a^{b}|gamma '(t)|, dt$. So we get $L=int _0^{3/8} |frac d {dt} e^{-2pi i t}| , dt=int _0^{3/8}2pi , dt=3pi /4$.
answered Dec 13 '18 at 12:18
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
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oldest
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active
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$begingroup$
$gamma$ is the part of the unit circle from $1$ to $-i$ and $|z|=1$ on the circle. So $M=1$. To find $L$ use the formula $L=int_a^{b}|gamma '(t)|, dt$. So we get $L=int _0^{3/8} |frac d {dt} e^{-2pi i t}| , dt=int _0^{3/8}2pi , dt=3pi /4$.
answered Dec 13 '18 at 12:18
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
add a comment |
$begingroup$
$gamma$ is the part of the unit circle from $1$ to $-i$ and $|z|=1$ on the circle. So $M=1$. To find $L$ use the formula $L=int_a^{b}|gamma '(t)|, dt$. So we get $L=int _0^{3/8} |frac d {dt} e^{-2pi i t}| , dt=int _0^{3/8}2pi , dt=3pi /4$.
answered Dec 13 '18 at 12:18
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
add a comment |
$begingroup$
$gamma$ is the part of the unit circle from $1$ to $-i$ and $|z|=1$ on the circle. So $M=1$. To find $L$ use the formula $L=int_a^{b}|gamma '(t)|, dt$. So we get $L=int _0^{3/8} |frac d {dt} e^{-2pi i t}| , dt=int _0^{3/8}2pi , dt=3pi /4$.
answered Dec 13 '18 at 12:18
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
$gamma$ is the part of the unit circle from $1$ to $-i$ and $|z|=1$ on the circle. So $M=1$. To find $L$ use the formula $L=int_a^{b}|gamma '(t)|, dt$. So we get $L=int _0^{3/8} |frac d {dt} e^{-2pi i t}| , dt=int _0^{3/8}2pi , dt=3pi /4$.
answered Dec 13 '18 at 12:18
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Dec 13 '18 at 12:18
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Dec 13 '18 at 12:18
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Dec 13 '18 at 12:18
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
63.7k42463
add a comment |
add a comment |
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