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Thread: Behavior as x,y -> infty

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    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?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fifthrapiers View Post
    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.

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

    If y goes to pos/neg infty, the function will oscillate.
    that's what i say
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    Quote Originally Posted by fifthrapiers View Post
    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.
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