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Deriving inverse tans from complex product?

  1. Apr 15, 2007 #1
    I am supposed to derive [itex]4\,\arctan \left( 1/5 \right) -\arctan \left( {\frac {1}{239}}
    \right) [/itex] from the complex product of [itex]\left( 1+i \right) \left( 5-i \right) ^{4}[/itex] I do see how the argument of product of the complex expression is equal to pi/4 - 4 arctan(1/5) but I am totally lost. SO how am I supposed to approach this? Thanks heaps in advance.
     
  2. jcsd
  3. Apr 15, 2007 #2

    Gib Z

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    Well You have to equate to long way and the short way :D

    You already established that the argument to the complex product is [itex]\frac{\pi}{4} -4\arctan \frac{1}{5}[/itex]

    What you had to do next is have the pleasure of EXPANDING YOUR PRODUCT. Binomial theorem is just too much of a hassle this time. Write the product as
    [tex](5-i)^{2^2}\cdot(1+i)[/tex]. So just square 5-i, 2 times, and multiply that by 1+i.

    FUN FUN, I did it for you :P You get : 956-4i . The argument of this is arctan (-4/956)= - arctan (1/239) . Isn't that interesting?
     
  4. Apr 16, 2007 #3
    And here I thought I'd never have to multiply the hard way after learning the polar form. Thanks heaps again =)
     
  5. Apr 16, 2007 #4

    Curious3141

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