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Homework Help: Kpi or k2pi?

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data

    This problem is probably quite simple for people who have covered this topic but the answer for part A is n=4k where k is any integer.

    Part B answer is n= 2+4k

    Look below for the working out and i will tell you what i dont understand

    2. Relevant equations
    de moivres theorem

    3. The attempt at a solution

    polar form is ([itex]\sqrt{2}[/itex] cis [itex]\pi[/itex] /4)^n

    so [itex]\sqrt{2}[/itex]^n [cos(n[itex]\pi[/itex] /4)] +isin(n[itex]\pi[/itex] /4)] (demoivres theorem)

    When Sin (n[itex]\pi[/itex] /4) = 0 the equation is real

    n[itex]\pi[/itex] /4 = 0 +k2[itex]\pi[/itex] (For some reason this should be +k[itex]\pi[/itex] which is why i dont get the answer.

    n=8k (solving for n)

    Part b is prety much the same thing but cos (n[itex]\pi[/itex] /4) The question that i am asking is why is it +k[itex]\pi[/itex] and not +k2[itex]\pi[/itex] as that is usually the period for sin and cosine but [itex]\pi[/itex] is the period for tan i know that.
  2. jcsd
  3. Sep 30, 2011 #2
    Try to find out the general solution for:-
  4. Sep 30, 2011 #3


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    What Pranav-Arora suggests is correct. All that's required to make (z+i)n real is for its imaginary part to be zero. It doesn't matter whether the real part is positive of if it's negative.

    Similarly, to make it purely imaginary, its real part must be zero.
  5. Sep 30, 2011 #4
    Even though the period is 2[itex]\pi[/itex], sinx = 0 at both 0 and at [itex]\pi[/itex].
  6. Sep 30, 2011 #5
    so are you suggesting that both answers are correct whether i write both pi and 2pi? as sin = 0 at both, how about cos its zero at pi/2 and 3pi/2 but in the book it sais pi aswell
  7. Sep 30, 2011 #6
    Okay ive figured out out now thankyou very much! :))
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