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Math Help - Behavior as x,y -> infty

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    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?
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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 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.

    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 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.
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