Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 years ago

Calculate the following: I. log(1+i√3) II. Log((e^2)i)?

Complex analysis

4 Answers

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  • 7 years ago
    Best Answer

    I.

    1+i√3 in polar form is 2 cis(π/3), which can be written 2e^(iπ/3)

    So ln(1+i√3) = ln(2e^(iπ/3))

    ln(2) + ln(e^(iπ/3))

    ln(2) + iπ/3

    ===

    If you account for rotations about the polar axis then there are in fact infinitely many solutions of the form ln(2) + iπ(2n + 1/3).

    ===

    @Admire, above, I gave you an up thumb. Why? Because you gave a valid albeit naïve answer. And I respect your humble edit.

  • 3 years ago

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  • Admire
    Lv 7
    7 years ago

    log of complex numbers do not exist, so I'd say

    I log(1 + 0) = log 1 = 0

    II log e^2 + log i

    2loge + 0 = 2loge

    Edit

    Seems others know a lot about this, so ignore my answer.

  • Anonymous
    7 years ago

    I. Log(1+i(3)^(1/2))

    =Log(2) + Log(1/2 + i(3)^(1/2))

    =Log(2) + Log(cos(pi/3) + isin(pi/3))

    =Log(2) + Log(e^(ipi/3))

    =Log(2) + ipi/3

    II. Log(e^2*i)

    =Log(e^2) + Log(i)

    =2 + Log(e^(ipi/2))

    =2 + ipi/2

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