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Let $g(z) = \exp(z^2)$ and $L$ a ray starting at the origin. Determine those $L$ along which $g$ has a limit (finite or infinite) as $|z|$ tends to infinity and $z ∈ L$. Find the value of the limit also. Please I need help.

Noble Mushtak
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  • Hint: $L$ can be parametrized by $z = re^{i\theta}$ where $\theta$ is fixed and $0 \le r < \infty$. – mrf Jan 24 '16 at 23:32
  • The limit is $0$ if $L$ is inside the "North" or "South" quadrants, $\infty$ if inside the "East" or "West" and doesn't exist if $L$ part of their boundary. – A.S. Jan 24 '16 at 23:33
  • @A.S. That's not quite true. (It looks like you forgot the square.) – mrf Jan 24 '16 at 23:34
  • @mrf - indeed. corrected. – A.S. Jan 24 '16 at 23:37
  • @A.S. The complex plane doesn't have a "north" quadrant. Its quadrants are "northeast" (Quadrant I), "northwest" (Quadrant II), "southwest" (Quadrant III) and "southeast" (Quadrant IV). Can you please make what you mean by "north," "south," "east," and "west" more specific? – Noble Mushtak Jan 24 '16 at 23:40
  • @Noble If you can figure out what NW and NE quadrants are, deducing what N quadrant stands for is pretty easy. The name of the quadrant is derived from the direction of its bisector. – A.S. Jan 24 '16 at 23:48
  • @A.S. There are two quadrants that I would classify as "north": Quadrants I and II. Likewise, there are two "south" quadrants, Quadrants III and IV, two "west" quadrants, Quadrants II and III, and two "east" quadrants, Quadrant I and IV. How am I supposed to know which is which? – Noble Mushtak Jan 24 '16 at 23:50
  • @Noble The name of the quadrant is derived from the direction of its bisector. There is only one North quadrant. Quarants I and II are in the North half-plane but are not North quadrants. – A.S. Jan 24 '16 at 23:54
  • Oh...are you splitting the complex plane into quadrants based off the lines containing $q+qi$ for $q \in \mathbb{R}$ and $q-qi$ for $q \in \mathbb{R}$? If so, I understand what you're talking about now. – Noble Mushtak Jan 24 '16 at 23:56
  • HINT: $$\begin{align} \left|e^{z^2}\right|&=\left|e^{R^2\cos(2\phi)}e^{iR^2\sin(2\phi)}\right|\\ &=e^{R^2\cos(2\phi)} \end{align}$$ – Mark Viola Jan 25 '16 at 00:00
  • @Noble Yes. I'm using quadrant as in the original "one quarter of a circle" sense which is more general than specific I,II,III,IV quadrants. – A.S. Jan 25 '16 at 00:04

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