For which $z$ goes $f(z) = e^{iz^2} rightarrow 0$ if $|z| rightarrow infty$?












2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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


















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















2












2








2


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    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












1 Answer
1






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oldest

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1












$begingroup$

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) $






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    1












    $begingroup$

    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) $






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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) $






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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) $






        share|cite|improve this answer









        $endgroup$



        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) $







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 11:15









        DylanDylan

        14.4k31127




        14.4k31127






























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