# Behavior as x,y -> infty

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

This is my answer:

We have $\displaystyle e^{x}(\cos{y} + i\sin{y})$. 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 $\displaystyle e^{x + iy}$ as $\displaystyle x \to \pm \infty$ and of $\displaystyle e^{x+iy}$ as $\displaystyle y \to \pm \infty$

This is my answer:

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

well, it depends on the value of $\displaystyle y$ (consider when $\displaystyle y = \pi$ 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
• Feb 5th 2008, 07:36 PM
ThePerfectHacker
Quote:

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

This is my answer:

We have $\displaystyle e^{x}(\cos{y} + i\sin{y})$. 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 $\displaystyle f(z) \to \infty$ not when $\displaystyle z\to \infty$ (because that is ambigous) but when $\displaystyle |z|\to \infty$. Hence the $\displaystyle |z|\to \infty$ does not exists because different path give different results.