For which $z$ goes $f(z) = e^{iz^2} rightarrow 0$ if $|z| rightarrow infty$?
$begingroup$
Consider the function:
$$f(z) = e^{iz^2}.$$
For which values of $z$ goes $f(z) rightarrow 0$ if $|z| rightarrow infty$.
I've read that it is possible if:
$$0 < arg(z) leq frac{pi}{4}.$$
I've started by writing $f(z)$ as:
$$e^{iz^2} = e^{i|z|^2e^{i2theta}}.$$
We can then write the power as:
$$i|z|^2e^{i2theta} = i|z|^2cos(2theta) - |z|^2sin(2theta).$$
If $f(z)$ needs to go to zero, then the power of the exponent needs to be negative so:
$$sin(2theta)> 0 $$
$$cos(2theta)<0 $$
But then I get that $theta geq frac{pi}{4}$
Am I missing something or is there a better way to proof this.
complex-analysis
$endgroup$
|
show 4 more comments
$begingroup$
Consider the function:
$$f(z) = e^{iz^2}.$$
For which values of $z$ goes $f(z) rightarrow 0$ if $|z| rightarrow infty$.
I've read that it is possible if:
$$0 < arg(z) leq frac{pi}{4}.$$
I've started by writing $f(z)$ as:
$$e^{iz^2} = e^{i|z|^2e^{i2theta}}.$$
We can then write the power as:
$$i|z|^2e^{i2theta} = i|z|^2cos(2theta) - |z|^2sin(2theta).$$
If $f(z)$ needs to go to zero, then the power of the exponent needs to be negative so:
$$sin(2theta)> 0 $$
$$cos(2theta)<0 $$
But then I get that $theta geq frac{pi}{4}$
Am I missing something or is there a better way to proof this.
complex-analysis
$endgroup$
$begingroup$
Why the condition $cos(2theta)<0$? Which source for the solution?
$endgroup$
– Did
Dec 21 '18 at 17:22
$begingroup$
if $cos(2theta) < 0$ then $e^{i|z|^2 cos(2theta}$ has a negative power so if $|z| rightarrow infty$ then the exponent will go to zero. And the source is mathematical methods for physics and engineers from Riley.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:25
$begingroup$
"then the exponent will go to zero" Sorry but you must be MUCH more precise here. Which exponent goes to zero?
$endgroup$
– Did
Dec 21 '18 at 17:27
$begingroup$
Sorry for the miscommunication, but I mean $e^{i|z|^2cos(2theta)} rightarrow 0$ if $cos(2theta) < 0$.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:28
$begingroup$
And this is not true since, for every real $|z|$ and $theta$, $$|e^{i|z|^2cos(2theta)}|=1$$
$endgroup$
– Did
Dec 21 '18 at 17:30
|
show 4 more comments
$begingroup$
Consider the function:
$$f(z) = e^{iz^2}.$$
For which values of $z$ goes $f(z) rightarrow 0$ if $|z| rightarrow infty$.
I've read that it is possible if:
$$0 < arg(z) leq frac{pi}{4}.$$
I've started by writing $f(z)$ as:
$$e^{iz^2} = e^{i|z|^2e^{i2theta}}.$$
We can then write the power as:
$$i|z|^2e^{i2theta} = i|z|^2cos(2theta) - |z|^2sin(2theta).$$
If $f(z)$ needs to go to zero, then the power of the exponent needs to be negative so:
$$sin(2theta)> 0 $$
$$cos(2theta)<0 $$
But then I get that $theta geq frac{pi}{4}$
Am I missing something or is there a better way to proof this.
complex-analysis
$endgroup$
Consider the function:
$$f(z) = e^{iz^2}.$$
For which values of $z$ goes $f(z) rightarrow 0$ if $|z| rightarrow infty$.
I've read that it is possible if:
$$0 < arg(z) leq frac{pi}{4}.$$
I've started by writing $f(z)$ as:
$$e^{iz^2} = e^{i|z|^2e^{i2theta}}.$$
We can then write the power as:
$$i|z|^2e^{i2theta} = i|z|^2cos(2theta) - |z|^2sin(2theta).$$
If $f(z)$ needs to go to zero, then the power of the exponent needs to be negative so:
$$sin(2theta)> 0 $$
$$cos(2theta)<0 $$
But then I get that $theta geq frac{pi}{4}$
Am I missing something or is there a better way to proof this.
complex-analysis
complex-analysis
asked Dec 21 '18 at 16:59
Belgium_PhysicsBelgium_Physics
322110
322110
$begingroup$
Why the condition $cos(2theta)<0$? Which source for the solution?
$endgroup$
– Did
Dec 21 '18 at 17:22
$begingroup$
if $cos(2theta) < 0$ then $e^{i|z|^2 cos(2theta}$ has a negative power so if $|z| rightarrow infty$ then the exponent will go to zero. And the source is mathematical methods for physics and engineers from Riley.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:25
$begingroup$
"then the exponent will go to zero" Sorry but you must be MUCH more precise here. Which exponent goes to zero?
$endgroup$
– Did
Dec 21 '18 at 17:27
$begingroup$
Sorry for the miscommunication, but I mean $e^{i|z|^2cos(2theta)} rightarrow 0$ if $cos(2theta) < 0$.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:28
$begingroup$
And this is not true since, for every real $|z|$ and $theta$, $$|e^{i|z|^2cos(2theta)}|=1$$
$endgroup$
– Did
Dec 21 '18 at 17:30
|
show 4 more comments
$begingroup$
Why the condition $cos(2theta)<0$? Which source for the solution?
$endgroup$
– Did
Dec 21 '18 at 17:22
$begingroup$
if $cos(2theta) < 0$ then $e^{i|z|^2 cos(2theta}$ has a negative power so if $|z| rightarrow infty$ then the exponent will go to zero. And the source is mathematical methods for physics and engineers from Riley.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:25
$begingroup$
"then the exponent will go to zero" Sorry but you must be MUCH more precise here. Which exponent goes to zero?
$endgroup$
– Did
Dec 21 '18 at 17:27
$begingroup$
Sorry for the miscommunication, but I mean $e^{i|z|^2cos(2theta)} rightarrow 0$ if $cos(2theta) < 0$.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:28
$begingroup$
And this is not true since, for every real $|z|$ and $theta$, $$|e^{i|z|^2cos(2theta)}|=1$$
$endgroup$
– Did
Dec 21 '18 at 17:30
$begingroup$
Why the condition $cos(2theta)<0$? Which source for the solution?
$endgroup$
– Did
Dec 21 '18 at 17:22
$begingroup$
Why the condition $cos(2theta)<0$? Which source for the solution?
$endgroup$
– Did
Dec 21 '18 at 17:22
$begingroup$
if $cos(2theta) < 0$ then $e^{i|z|^2 cos(2theta}$ has a negative power so if $|z| rightarrow infty$ then the exponent will go to zero. And the source is mathematical methods for physics and engineers from Riley.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:25
$begingroup$
if $cos(2theta) < 0$ then $e^{i|z|^2 cos(2theta}$ has a negative power so if $|z| rightarrow infty$ then the exponent will go to zero. And the source is mathematical methods for physics and engineers from Riley.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:25
$begingroup$
"then the exponent will go to zero" Sorry but you must be MUCH more precise here. Which exponent goes to zero?
$endgroup$
– Did
Dec 21 '18 at 17:27
$begingroup$
"then the exponent will go to zero" Sorry but you must be MUCH more precise here. Which exponent goes to zero?
$endgroup$
– Did
Dec 21 '18 at 17:27
$begingroup$
Sorry for the miscommunication, but I mean $e^{i|z|^2cos(2theta)} rightarrow 0$ if $cos(2theta) < 0$.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:28
$begingroup$
Sorry for the miscommunication, but I mean $e^{i|z|^2cos(2theta)} rightarrow 0$ if $cos(2theta) < 0$.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:28
$begingroup$
And this is not true since, for every real $|z|$ and $theta$, $$|e^{i|z|^2cos(2theta)}|=1$$
$endgroup$
– Did
Dec 21 '18 at 17:30
$begingroup$
And this is not true since, for every real $|z|$ and $theta$, $$|e^{i|z|^2cos(2theta)}|=1$$
$endgroup$
– Did
Dec 21 '18 at 17:30
|
show 4 more comments
1 Answer
1
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$begingroup$
Let $z = x + iy$, then
$$ f(z) = e^{i(x+iy)^2} = e^{i(x^2-y^2+2ixy)} = e^{-2xy}e^{i(x^2-y^2)} $$
$|f| = e^{-2xy} to 0$ if $xy > 0$, which means $z$ must be in the first or third quadrant, or $arg(z) in (0,pi/2) cup (-pi/2,-pi) $
$endgroup$
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1 Answer
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1 Answer
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oldest
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$begingroup$
Let $z = x + iy$, then
$$ f(z) = e^{i(x+iy)^2} = e^{i(x^2-y^2+2ixy)} = e^{-2xy}e^{i(x^2-y^2)} $$
$|f| = e^{-2xy} to 0$ if $xy > 0$, which means $z$ must be in the first or third quadrant, or $arg(z) in (0,pi/2) cup (-pi/2,-pi) $
$endgroup$
add a comment |
$begingroup$
Let $z = x + iy$, then
$$ f(z) = e^{i(x+iy)^2} = e^{i(x^2-y^2+2ixy)} = e^{-2xy}e^{i(x^2-y^2)} $$
$|f| = e^{-2xy} to 0$ if $xy > 0$, which means $z$ must be in the first or third quadrant, or $arg(z) in (0,pi/2) cup (-pi/2,-pi) $
$endgroup$
add a comment |
$begingroup$
Let $z = x + iy$, then
$$ f(z) = e^{i(x+iy)^2} = e^{i(x^2-y^2+2ixy)} = e^{-2xy}e^{i(x^2-y^2)} $$
$|f| = e^{-2xy} to 0$ if $xy > 0$, which means $z$ must be in the first or third quadrant, or $arg(z) in (0,pi/2) cup (-pi/2,-pi) $
$endgroup$
Let $z = x + iy$, then
$$ f(z) = e^{i(x+iy)^2} = e^{i(x^2-y^2+2ixy)} = e^{-2xy}e^{i(x^2-y^2)} $$
$|f| = e^{-2xy} to 0$ if $xy > 0$, which means $z$ must be in the first or third quadrant, or $arg(z) in (0,pi/2) cup (-pi/2,-pi) $
answered Dec 23 '18 at 11:15
DylanDylan
14.4k31127
14.4k31127
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$begingroup$
Why the condition $cos(2theta)<0$? Which source for the solution?
$endgroup$
– Did
Dec 21 '18 at 17:22
$begingroup$
if $cos(2theta) < 0$ then $e^{i|z|^2 cos(2theta}$ has a negative power so if $|z| rightarrow infty$ then the exponent will go to zero. And the source is mathematical methods for physics and engineers from Riley.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:25
$begingroup$
"then the exponent will go to zero" Sorry but you must be MUCH more precise here. Which exponent goes to zero?
$endgroup$
– Did
Dec 21 '18 at 17:27
$begingroup$
Sorry for the miscommunication, but I mean $e^{i|z|^2cos(2theta)} rightarrow 0$ if $cos(2theta) < 0$.
$endgroup$
– Belgium_Physics
Dec 21 '18 at 17:28
$begingroup$
And this is not true since, for every real $|z|$ and $theta$, $$|e^{i|z|^2cos(2theta)}|=1$$
$endgroup$
– Did
Dec 21 '18 at 17:30