Behavior as x,y -> infty

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January 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 $y \to \pm \infty$

This is my answer:

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?
January 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 $y \to \pm \infty$

This is my answer:

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.

Quote:

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

yup

Quote:

If y goes to pos/neg infty, the function will oscillate.

that's what i say
February 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 $y \to \pm \infty$

This is my answer:

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?

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.