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Basic diff eq math questions

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data
    I have some trouble understanding some of my math homework maybe somebody can help me out?

    1) find all the values of
    ln(e)
    (-1)^i

    I know that i am going to have to use eulers formula in some way i believe but im not really sure what the question is asking, what does it mean all the values?

    2)let z= x+iy, where both x and y are real. find the real and imaginary parts for e^(1/z).
    i figure i would have e^(1/x+iy) but where do i go from there?

    3) rewrite in the polar form
    (√(z))^(1/n), now this i dont really understand. I can really easily change something like 4 +6i to polar form but this i dont get.
     
  2. jcsd
  3. Jan 21, 2012 #2
    When things say all values they generally mean the 'general solution'
    Like [itex]ArcCos(1)=2n \pi [/itex]

    For part two, you could try multiplying 1/z by 1 in such a way that you end up with a real denominator (you should know how to do this)

    For part three you just need to know about how exponents work, what happens when you take the square root of a^x, what happens when you take the nth power of a^x?
     
  4. Jan 22, 2012 #3
    still dont really understand part 1,

    thanks i got part 2, was just a stupid mistake on my part
    and for part 3 i know how exponents work in the sense that i am taking the nth root of z but how does one convert that to a polar form.
     
  5. Jan 22, 2012 #4
    The exponential function is periodic, with period [itex]2\pi i[/itex] (that is, [itex]e^z=e^{z+2\pi i}[/itex] for every [itex]z\in\mathbb{C}[/itex]), and the logarithm is defined as the inverse of the exponential. Since the exponential is not one-to-one, the logarithm of a number is not uniquely determined; it's not an actual function on [itex]\mathbb{C}[/itex]. The best you can do to find [itex]\log(z)[/itex] is to give a set of complex numbers (differing by integer multiples of [itex]2\pi i[/itex]) whose exponential is [itex]z[/itex].

    Likewise, since [itex]a^z[/itex] is defined as [itex]e^{z\log(a)}[/itex], it is also multivalued.
     
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