# Behavior as x,y -> infty

• Jan 29th 2008, 06:33 PM
fifthrapiers
Behavior as x,y -> infty
Describe the behavior of $e^{x + iy}$ as $x \to \pm \infty$ and of $e^{x+iy}$ as $y \to \pm \infty$

We have $e^{x}(\cos{y} + i\sin{y})$. This implies that if x -> infinity, the function will go to infinity. And, if x -> neg. infty, the function will go to 0. If y goes to pos/neg infty, the function will oscillate.

Thoughts?
• Jan 29th 2008, 06:56 PM
Jhevon
Quote:

Originally Posted by fifthrapiers
Describe the behavior of $e^{x + iy}$ as $x \to \pm \infty$ and of $e^{x+iy}$ as $y \to \pm \infty$

We have $e^{x}(\cos{y} + i\sin{y})$. This implies that if x -> infinity, the function will go to infinity.

well, it depends on the value of $y$ (consider when $y = \pi$ for instance). but in general, i guess you could say so. not exactly sure how you define a complex number going to infinity.

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And, if x -> neg. infty, the function will go to 0.
yup

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If y goes to pos/neg infty, the function will oscillate.
that's what i say
• Feb 5th 2008, 07:36 PM
ThePerfectHacker
Quote:

Originally Posted by fifthrapiers
Describe the behavior of $e^{x + iy}$ as $x \to \pm \infty$ and of $e^{x+iy}$ as $y \to \pm \infty$

We have $e^{x}(\cos{y} + i\sin{y})$. This implies that if x -> infinity, the function will go to infinity. And, if x -> neg. infty, the function will go to 0. If y goes to pos/neg infty, the function will oscillate.
In the complex plane there are many directions to infinity. The standard definition is that we say $f(z) \to \infty$ not when $z\to \infty$ (because that is ambigous) but when $|z|\to \infty$. Hence the $|z|\to \infty$ does not exists because different path give different results.