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

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.

• Why the condition $\cos(2\theta)<0$? Which source for the solution? – Did Dec 21 '18 at 17:22
• 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. – Belgium_Physics Dec 21 '18 at 17:25
• "then the exponent will go to zero" Sorry but you must be MUCH more precise here. Which exponent goes to zero? – Did Dec 21 '18 at 17:27
• Sorry for the miscommunication, but I mean $e^{i|z|^2cos(2\theta)} \rightarrow 0$ if $cos(2\theta) < 0$. – Belgium_Physics Dec 21 '18 at 17:28
• And this is not true since, for every real $|z|$ and $\theta$, $$|e^{i|z|^2\cos(2\theta)}|=1$$ – 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 $$\arg(z) \in (0,\pi/2) \cup (-\pi/2,-\pi)$$