$|e^{z_1}-e^{z_2}|$ is less than $|z_1-z_2|$ if the real parts are non-positive
Is $|e^{z_1}-e^{z_2}| le|z_1-z_2|$ if the real parts of $z_1,z_2$ are non-positive?
I think yes, but what method do I proceed with? Do I proceed with the fact that modulus equals the number multiplied by its conjugate, or do I expand the complex numbers into real and imaginary parts. Any hints? Thanks beforehand.
calculus algebra-precalculus complex-numbers
asked Dec 6 at 17:06
vidyarthi
2,8021831
add a comment |
Is $|e^{z_1}-e^{z_2}| le|z_1-z_2|$ if the real parts of $z_1,z_2$ are non-positive?
I think yes, but what method do I proceed with? Do I proceed with the fact that modulus equals the number multiplied by its conjugate, or do I expand the complex numbers into real and imaginary parts. Any hints? Thanks beforehand.
calculus algebra-precalculus complex-numbers
asked Dec 6 at 17:06
vidyarthi
2,8021831
add a comment |
Is $|e^{z_1}-e^{z_2}| le|z_1-z_2|$ if the real parts of $z_1,z_2$ are non-positive?
I think yes, but what method do I proceed with? Do I proceed with the fact that modulus equals the number multiplied by its conjugate, or do I expand the complex numbers into real and imaginary parts. Any hints? Thanks beforehand.
calculus algebra-precalculus complex-numbers
asked Dec 6 at 17:06
vidyarthi
2,8021831
Is $|e^{z_1}-e^{z_2}| le|z_1-z_2|$ if the real parts of $z_1,z_2$ are non-positive?
I think yes, but what method do I proceed with? Do I proceed with the fact that modulus equals the number multiplied by its conjugate, or do I expand the complex numbers into real and imaginary parts. Any hints? Thanks beforehand.
calculus algebra-precalculus complex-numbers
calculus algebra-precalculus complex-numbers
asked Dec 6 at 17:06
vidyarthi
2,8021831
asked Dec 6 at 17:06
vidyarthi
2,8021831
edited Dec 6 at 17:27
asked Dec 6 at 17:06
vidyarthi
2,8021831
asked Dec 6 at 17:06
vidyarthi
2,8021831
asked Dec 6 at 17:06
vidyarthi
2,8021831
2,8021831
add a comment |
add a comment |
1 Answer
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Hint:
$f(z_1) - f(z_2) = int_C f'(z); dz $ where $C$ is the directed line segment from $z_2$ to $z_1$, so $|f(z_1) - f(z_2)| le |z_1 - z_2| sup_{z in C} |f'(z)|$.
answered Dec 6 at 17:24
Robert Israel
317k23206457
oh! ok, so its just the mean value theorem for complex numbers, i suppose?
– vidyarthi
Dec 6 at 17:26
This might be considered as related to the MVT. Note that the MVT itself is not true for complex numbers, i.e. there might not be any $z in C$ for which $f'(z) = (f(z_1) - f(z_2))/(z_1 - z_2)$.
– Robert Israel
Dec 6 at 17:36
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
$f(z_1) - f(z_2) = int_C f'(z); dz $ where $C$ is the directed line segment from $z_2$ to $z_1$, so $|f(z_1) - f(z_2)| le |z_1 - z_2| sup_{z in C} |f'(z)|$.
answered Dec 6 at 17:24
Robert Israel
317k23206457
oh! ok, so its just the mean value theorem for complex numbers, i suppose?
– vidyarthi
Dec 6 at 17:26
This might be considered as related to the MVT. Note that the MVT itself is not true for complex numbers, i.e. there might not be any $z in C$ for which $f'(z) = (f(z_1) - f(z_2))/(z_1 - z_2)$.
– Robert Israel
Dec 6 at 17:36
add a comment |
Hint:
$f(z_1) - f(z_2) = int_C f'(z); dz $ where $C$ is the directed line segment from $z_2$ to $z_1$, so $|f(z_1) - f(z_2)| le |z_1 - z_2| sup_{z in C} |f'(z)|$.
answered Dec 6 at 17:24
Robert Israel
317k23206457
oh! ok, so its just the mean value theorem for complex numbers, i suppose?
– vidyarthi
Dec 6 at 17:26
This might be considered as related to the MVT. Note that the MVT itself is not true for complex numbers, i.e. there might not be any $z in C$ for which $f'(z) = (f(z_1) - f(z_2))/(z_1 - z_2)$.
– Robert Israel
Dec 6 at 17:36
add a comment |
Hint:
$f(z_1) - f(z_2) = int_C f'(z); dz $ where $C$ is the directed line segment from $z_2$ to $z_1$, so $|f(z_1) - f(z_2)| le |z_1 - z_2| sup_{z in C} |f'(z)|$.
answered Dec 6 at 17:24
Robert Israel
317k23206457
Hint:
$f(z_1) - f(z_2) = int_C f'(z); dz $ where $C$ is the directed line segment from $z_2$ to $z_1$, so $|f(z_1) - f(z_2)| le |z_1 - z_2| sup_{z in C} |f'(z)|$.
answered Dec 6 at 17:24
Robert Israel
317k23206457
answered Dec 6 at 17:24
Robert Israel
317k23206457
answered Dec 6 at 17:24
Robert Israel
317k23206457
answered Dec 6 at 17:24
Robert Israel
317k23206457
317k23206457
oh! ok, so its just the mean value theorem for complex numbers, i suppose?
– vidyarthi
Dec 6 at 17:26
This might be considered as related to the MVT. Note that the MVT itself is not true for complex numbers, i.e. there might not be any $z in C$ for which $f'(z) = (f(z_1) - f(z_2))/(z_1 - z_2)$.
– Robert Israel
Dec 6 at 17:36
add a comment |
oh! ok, so its just the mean value theorem for complex numbers, i suppose?
– vidyarthi
Dec 6 at 17:26
This might be considered as related to the MVT. Note that the MVT itself is not true for complex numbers, i.e. there might not be any $z in C$ for which $f'(z) = (f(z_1) - f(z_2))/(z_1 - z_2)$.
– Robert Israel
Dec 6 at 17:36
oh! ok, so its just the mean value theorem for complex numbers, i suppose?
– vidyarthi
Dec 6 at 17:26
oh! ok, so its just the mean value theorem for complex numbers, i suppose?
– vidyarthi
Dec 6 at 17:26
This might be considered as related to the MVT. Note that the MVT itself is not true for complex numbers, i.e. there might not be any $z in C$ for which $f'(z) = (f(z_1) - f(z_2))/(z_1 - z_2)$.
– Robert Israel
Dec 6 at 17:36
This might be considered as related to the MVT. Note that the MVT itself is not true for complex numbers, i.e. there might not be any $z in C$ for which $f'(z) = (f(z_1) - f(z_2))/(z_1 - z_2)$.
– Robert Israel
Dec 6 at 17:36
add a comment |
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