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Complex Powers

  1. Jul 1, 2003 #1
    How do we express complex powers of numbers (e.g. 21+i) in the form a+bi, or some other standard form of representation for complex numbers?
     
  2. jcsd
  3. Jul 1, 2003 #2

    HallsofIvy

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    First, of course, 21+i= 2*2i so the question is really about 2i (or, more generally, abi).

    Specfically, look at eix.

    It is possible to show (using Taylor's series) that

    e^(ix)= cos(x)+ i sin(x).

    a^(bi)= e^(ln(a^(bi))= e^(bi*ln(a))= cos(b ln(a))+ i sin(b ln(a))
    = cos(ln(a^b))+ i sin(ln(a^b))

    For your particular case, 2^i= cos(ln(2))+ i sin(ln(2))
    = 0.769+ 0.639 i.

    2^(1+i)= 2(0.769+ 0.639i)= 0.1538+ 1.278 i.
     
  4. Jul 1, 2003 #3
    Now why didn't I see that? Oh well, thanks for pointing it out. :smile:
     
  5. Jul 1, 2003 #4

    chroot

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    You're no doubt familiar with Euler's expression

    exp(i x) = cos(x) + i sin(x)

    You're probably also familiar that logarithms can be expressed in any base you'd like, like this:

    loga x = ( logb x ) / ( logb] a )

    For example, if your calculator has only log base 10, and you want to compute log2 16, you could enter

    log10 16 / log10 2

    We can put these facts together to good use.

    To start with, let's try a simple one: express 2i in the a + bi form. We can express 2i as a power of e by solving this equation:

    2i = ex
    i ln 2 = x

    We've just used the logarithm rule I described above in "reverse." So we've just changed the problem to expressing exp(i ln 2) in a + bi form. Now we can just apply Euler's identity, and we get

    exp(i ln 2) = cos(ln 2) + i sin(ln 2).

    Thus 2i = cos(ln 2) + i sin(ln 2), as we wished to find.

    Now let's try 21 + i. I'm going to skip all the fanfare and just show the steps.

    21+i = ex
    (1+i) ln 2 = x

    e(1+i) ln 2 = 21+i
    eln 2 + i ln 2
    eln 2 ei ln 2
    2 ei ln 2
    2 [ cos(ln 2) + i sin(ln 2) ]
    2 cos(ln 2) + 2 i sin(ln 2)

    Hope this helps.

    - Warren
     
    Last edited: Jul 1, 2003
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