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: $$e^{iz^2} = e^{i|z|^2e^{i2\theta}}.$$ We can then write the power as: $$i|z|^2e^{i2\theta} = i|z|^2cos(2\theta) - |z|^2sin(2\theta).$$ If $f(z)$ needs to go to zero, then the power of the exponent needs to be negative so: $$sin(2\theta)> 0 $$ $$cos(2\theta)<0 $$

But then I get that $\theta \geq \frac{\pi}{4}$

Am I missing something or is there a better way to proof this.

  • $\begingroup$ Why the condition $\cos(2\theta)<0$? Which source for the solution? $\endgroup$ – Did Dec 21 '18 at 17:22
  • $\begingroup$ if $cos(2\theta) < 0$ then $e^{i|z|^2 cos(2\theta}$ 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 $e^{i|z|^2cos(2\theta)} \rightarrow 0$ if $cos(2\theta) < 0$. $\endgroup$ – Belgium_Physics Dec 21 '18 at 17:28
  • $\begingroup$ And this is not true since, for every real $|z|$ and $\theta$, $$|e^{i|z|^2\cos(2\theta)}|=1$$ $\endgroup$ – Did Dec 21 '18 at 17:30

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) $


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