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Finding polar and cartesian form for this power

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    ((-1+i)/(√2))^1002
    find polar and cartesian form

    2. Relevant equations



    3. The attempt at a solution


    So I started by finding |z|=1

    and Arg(z)= arctan (-1) = 5pi/6

    so 1^(1002)*e^(i*5pi/6)*1002 =1*e^(i*835pi)

    but thats as far as I got because the answer says -i, but to get that term I need 1*e^-ipi/2

    any thoughts??
     
  2. jcsd
  3. Apr 11, 2012 #2

    HallsofIvy

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    Your main problem is that [itex]arctan(-1)[/itex] is NOT [itex]5\pi/6[/itex]!

    Calculate that again.
     
  4. Apr 11, 2012 #3
    I was assuming since the original eqn has a negative real part then I would add Pi/2 since the angle is measured from 0.
    Although Im getting 3pi/4 now
     
  5. Apr 11, 2012 #4
    Well, yes (3pi)/4 is the angle you're looking for since this is in the second quadrant.

    You have got |z|, and now you have the angle.

    Put it into exponential form to the power of 1002, and simplify it to get the angle in terms of a principal argument.
     
  6. Apr 11, 2012 #5
    Ok so now I have 1^1002*(cos 3Pi/4+isin3Pi/4)

    = 1*(0)=0 how do they get the minus i?
     
  7. Apr 11, 2012 #6
    How did you get 0 for (cos 3Pi/4 + isin 3Pi/4) ?

    Also, you have forgotten to raise that to the power.
     
  8. Apr 12, 2012 #7
    isnt cos 3pi/4=-0.7...

    and sin 3pi/4=0.7...
     
  9. Apr 12, 2012 #8
    Yes, but you have a complex number there.

    -0.707 +0.707i

    You can't add them just like that.

    What of the 1002? You need to account for that too.
     
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