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Principal value

  1. May 31, 2014 #1
    Can someone please check if I got the answer right? I don't have the right answers for my exercise so I need someone to check it for me, please.

    Please find the principle value of (1+i)^-i

    My attempt:
    √2^(- i) = (e^ln√2)^(- i) = e^-i•ln√2
              = cos(- ln√2) + i•sin(- ln√2)
              = cos(ln√2) - i•sin(ln√2)

          (1 + i) = √2[(1/√2) + i(1/√2)]
              = √2[cos(π/4) + isin(π/4)]
              = √2•e^i(π/4)

    (1 + i)^(- i) = [√2•e^i(π/4)]^(- i)
           =[e^(π/4)] [√2^(- i)]
           =[e^(π/4)][cos(ln√2) - i•sin(ln√2)]

    Therefore (1 + i)^(- i) = [e^(π/4)][cos(ln√2) - i•sin(ln√2)]
  2. jcsd
  3. Jun 1, 2014 #2


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    It's not clear what these calculations represent:

    You seem to have trouble expressing (1+i) in polar form and in applying Euler's formula correctly.


    z = (x+iy) = r(cos θ + i sin θ) = r e[itex]^{i θ}[/itex]


    r = [itex]\sqrt{x^{2}+y^{2}}[/itex]


    θ = arctan (y/x)
  4. Jun 1, 2014 #3
    But what's wrong? r=√2 and θ=∏/4.
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