Complex number inequality, $|e^{z_1}-e^{z_2}| leq |z_1 - z_2|$ if $Re(z_1),Re(z_2) leq 0$
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I'm trying to show the complex inequality $|e^{z_1}-e^{z_2}| leq |z_1 - z_2|$ holds if $Re(z_1),Re(z_2) leq 0$. It seems intuitively obvious but I haven't been able to find something that works. Would appreciate a hint.
complex-analysis inequality complex-numbers
asked Aug 27 '13 at 17:08
rowrow
8415
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add a comment |
$begingroup$
I'm trying to show the complex inequality $|e^{z_1}-e^{z_2}| leq |z_1 - z_2|$ holds if $Re(z_1),Re(z_2) leq 0$. It seems intuitively obvious but I haven't been able to find something that works. Would appreciate a hint.
complex-analysis inequality complex-numbers
asked Aug 27 '13 at 17:08
rowrow
8415
$endgroup$
$begingroup$
You should probably start with proving $|e^z - 1| le |z|$ for $Re(z) le 0$.
$endgroup$
– Shitikanth
Aug 27 '13 at 17:33
add a comment |
$begingroup$
I'm trying to show the complex inequality $|e^{z_1}-e^{z_2}| leq |z_1 - z_2|$ holds if $Re(z_1),Re(z_2) leq 0$. It seems intuitively obvious but I haven't been able to find something that works. Would appreciate a hint.
complex-analysis inequality complex-numbers
asked Aug 27 '13 at 17:08
rowrow
8415
$endgroup$
I'm trying to show the complex inequality $|e^{z_1}-e^{z_2}| leq |z_1 - z_2|$ holds if $Re(z_1),Re(z_2) leq 0$. It seems intuitively obvious but I haven't been able to find something that works. Would appreciate a hint.
complex-analysis inequality complex-numbers
complex-analysis inequality complex-numbers
asked Aug 27 '13 at 17:08
rowrow
8415
asked Aug 27 '13 at 17:08
rowrow
8415
asked Aug 27 '13 at 17:08
rowrow
8415
asked Aug 27 '13 at 17:08
rowrow
8415
asked Aug 27 '13 at 17:08
rowrow
8415
8415
$begingroup$
You should probably start with proving $|e^z - 1| le |z|$ for $Re(z) le 0$.
$endgroup$
– Shitikanth
Aug 27 '13 at 17:33
add a comment |
$begingroup$
You should probably start with proving $|e^z - 1| le |z|$ for $Re(z) le 0$.
$endgroup$
– Shitikanth
Aug 27 '13 at 17:33
$begingroup$
You should probably start with proving $|e^z - 1| le |z|$ for $Re(z) le 0$.
$endgroup$
– Shitikanth
Aug 27 '13 at 17:33
$begingroup$
You should probably start with proving $|e^z - 1| le |z|$ for $Re(z) le 0$.
$endgroup$
– Shitikanth
Aug 27 '13 at 17:33
add a comment |
1 Answer
1
active
oldest
votes
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Let $gamma$ be a straight line from $z_2$ to $z_1$ (since the left half plane is convex, $operatorname{Re} z le 0$ for all $zingamma$). Then
$$ e^{z_1} - e^{z_2} = int_gamma e^z,dz, $$
so
$$lvert e^{z_1} - e^{z_2} rvert= leftlvert int_gamma e^z,dz rightrvertle ell(gamma) max_{zingamma} |e^z| le |z_1 - z_2| $$
since $|e^z| = |e^{x+iy}| = e^{x} le 1$ on $gamma$ because $xle 0$.
edited Nov 29 '18 at 20:36
mathstackuser
681112
answered Aug 27 '13 at 17:57
mrfmrf
37.4k54685
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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
$begingroup$
Let $gamma$ be a straight line from $z_2$ to $z_1$ (since the left half plane is convex, $operatorname{Re} z le 0$ for all $zingamma$). Then
$$ e^{z_1} - e^{z_2} = int_gamma e^z,dz, $$
so
$$lvert e^{z_1} - e^{z_2} rvert= leftlvert int_gamma e^z,dz rightrvertle ell(gamma) max_{zingamma} |e^z| le |z_1 - z_2| $$
since $|e^z| = |e^{x+iy}| = e^{x} le 1$ on $gamma$ because $xle 0$.
edited Nov 29 '18 at 20:36
mathstackuser
681112
answered Aug 27 '13 at 17:57
mrfmrf
37.4k54685
$endgroup$
add a comment |
$begingroup$
Let $gamma$ be a straight line from $z_2$ to $z_1$ (since the left half plane is convex, $operatorname{Re} z le 0$ for all $zingamma$). Then
$$ e^{z_1} - e^{z_2} = int_gamma e^z,dz, $$
so
$$lvert e^{z_1} - e^{z_2} rvert= leftlvert int_gamma e^z,dz rightrvertle ell(gamma) max_{zingamma} |e^z| le |z_1 - z_2| $$
since $|e^z| = |e^{x+iy}| = e^{x} le 1$ on $gamma$ because $xle 0$.
edited Nov 29 '18 at 20:36
mathstackuser
681112
answered Aug 27 '13 at 17:57
mrfmrf
37.4k54685
$endgroup$
add a comment |
$begingroup$
Let $gamma$ be a straight line from $z_2$ to $z_1$ (since the left half plane is convex, $operatorname{Re} z le 0$ for all $zingamma$). Then
$$ e^{z_1} - e^{z_2} = int_gamma e^z,dz, $$
so
$$lvert e^{z_1} - e^{z_2} rvert= leftlvert int_gamma e^z,dz rightrvertle ell(gamma) max_{zingamma} |e^z| le |z_1 - z_2| $$
since $|e^z| = |e^{x+iy}| = e^{x} le 1$ on $gamma$ because $xle 0$.
edited Nov 29 '18 at 20:36
mathstackuser
681112
answered Aug 27 '13 at 17:57
mrfmrf
37.4k54685
$endgroup$
Let $gamma$ be a straight line from $z_2$ to $z_1$ (since the left half plane is convex, $operatorname{Re} z le 0$ for all $zingamma$). Then
$$ e^{z_1} - e^{z_2} = int_gamma e^z,dz, $$
so
$$lvert e^{z_1} - e^{z_2} rvert= leftlvert int_gamma e^z,dz rightrvertle ell(gamma) max_{zingamma} |e^z| le |z_1 - z_2| $$
since $|e^z| = |e^{x+iy}| = e^{x} le 1$ on $gamma$ because $xle 0$.
edited Nov 29 '18 at 20:36
mathstackuser
681112
answered Aug 27 '13 at 17:57
mrfmrf
37.4k54685
edited Nov 29 '18 at 20:36
mathstackuser
681112
edited Nov 29 '18 at 20:36
mathstackuser
681112
edited Nov 29 '18 at 20:36
mathstackuser
681112
681112
answered Aug 27 '13 at 17:57
mrfmrf
37.4k54685
answered Aug 27 '13 at 17:57
mrfmrf
37.4k54685
answered Aug 27 '13 at 17:57
mrfmrf
37.4k54685
37.4k54685
add a comment |
add a comment |
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$begingroup$
You should probably start with proving $|e^z - 1| le |z|$ for $Re(z) le 0$.
$endgroup$
– Shitikanth
Aug 27 '13 at 17:33