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Complex numbers help

  1. Jan 12, 2008 #1
    how do I show that:

    |y + x|^2 = |y|^2 + |x|^2 + 2|yx|cos(a1-a2)

    where y = |y|exp(ia1)
    and
    where x = |x|exp(ia2)

    and how do I show that |exp(z)| = exp(Re(z)) where Re is the real part of an imaginary number z.

    thanks is advance
     
  2. jcsd
  3. Jan 13, 2008 #2
    Okay, your first question was to show that

    |y + x|^2 = |y|^2 + |x|^2 + 2|yx|cos(a1-a2)

    Using the definitions of y and x given. Start by writting down that

    y = |y|exp(ia1) = |y|(cos(a1)+isin(a1))
    x = |x|exp(ia2) = |x|(cos(a2)+isin(a2))

    Using the series definitions of the exponential, sine and cosine functions.

    You should now be able to see that

    |y + x|^2 = | |y|cos(a1) + |x|cos(a2) + i( |y|sin(a1) + |x|sin(a2) ) |^2

    = ( |y|cos(a1) + |x|cos(a2) )^2 + ( |y|sin(a1) + |x|sin(a2) )^2 (by getting rid of the modulus)
    = |y|^2cos^2(a1) + |y|^2sin^2(a1) + |x|^2cos^2(a2) + |x|^2sin^2(a1) + 2|xy|(cos(a1)cos(a2) + sin(a1)sin(a2))

    = |y|^2 + |x|^2 + 2|xy|cos( a1 - a2)

    Above i expanded the brackets, used the sin^2/cos^2 identity, the identity for the cosine of the difference of two angles. Hope this helped.

    For the second question:

    Let z = x + iy

    |exp(z)| = |exp(x + iy)| = |exp(x)*exp(iy)|

    = |exp(x)(cosy + isiny)|
    = |exp(x)cosy + iexp(x)siny|
    = sqrt(exp(2x)cos^2(y) + exp(2x)sin^2(y))
    = sqrt(exp(2x))
    = exp(x)
    = exp(Re(z))
     
  4. Jan 13, 2008 #3

    mathwonk

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    isn't that first question just the law of cosines? or are you saying you want to prove the law of cosines? i.e. is it allowed to use trig? if so there is nothing more to do.
     
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