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Expand (1-2i)^10 without the Binomial Expansion Theorem

  1. Dec 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Expand (1-2i)^10 without the Binomial Expansion Theorem



    I know I need to put this in polar form and then it's simple from there, however, I am simply having a difficult time finding the angle. Drawing the complex number as a vector in the complex plane I get a right triangle with sides equal to 1 and 2, so the magnitude of the vector is √5. I can take the inverse tangent of 2/1 but I don't get an angle that is "easy" to work with, i.e. something familiar like -pi/3, -pi/6, etc. My professor did make it clear those would be the only angles we should be familiar with for these types of problems.

    what am I missing?
     
  2. jcsd
  3. Dec 17, 2013 #2

    HallsofIvy

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    I would hate to call your professor a liar (perhaps mistaken would sound better?) but this is NOT an "easy" angle. The arctan of -2 is about -1.01 radians.
     
  4. Dec 17, 2013 #3

    D H

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    Sure it is. For any even integer n (e.g., 10), ##\cos(n\arctan(2))## is a rational, as is ##\sin(n\arctan(2))##.
     
  5. Dec 17, 2013 #4
    using that knowledge I get:

    √5*[cos(10arctan(2)) + i*(sin(10arctan(2))]

    I'm not sure how to express this is a+bi form (sorry, I perhaps should have mentioned that it needed to be expressed in that form).
     
  6. Dec 17, 2013 #5

    D H

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    You forget to raise √5 to the tenth power, and you have a sign error in your angle.

    After fixing those two issues, all you need to do is compute cos(10*arctan(2)) and sin(10*arctan(2)).

    Hint #1: What are cos(2*arctan(2)) and sin(2*arctan(2))?
    Hint #2: Use the fact that 10=2*5.
     
  7. Dec 17, 2013 #6
    I have now:

    (√5)10[cos(5*2arctan(2)) - isin(5*2arctan(2)]

    =(√5)10[cos(5*-0.6) - isin(5*-0.6)]

    =(√5)10[cos(3) - isin(3)]

    If this is correct i'm not sure how to proceed
     
  8. Dec 17, 2013 #7

    D H

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    You made a mistake going from step 1 to step 2. Use your trig identities.
     
  9. Dec 18, 2013 #8
    I've been trying to work this through for a while now and I just can't see any use for the trig identities (trig was always a weak point for me :confused: )

    haven't found any trig identities on the web containing inverse tangent. And arctan isn't the same as 1/tan.

    I'm sorry but I'm stumped.
     
  10. Dec 18, 2013 #9

    Office_Shredder

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    You typically don't want to evaluate arctan(2) when doing these things. Do you know how to evaluate sin(arctan(2)) by itself?
     
  11. Dec 18, 2013 #10

    FeDeX_LaTeX

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    Sure there are. You just need to play around with your identities a little. For example:

    ##\sin^{2}x + \cos^{2}x = 1 \implies \sin x = \sqrt{1 - \cos^{2}x} \implies \sin(\arccos x) = \sqrt{1 - x^2}##.

    Can you find a similar identity for ##\cos(k \arctan x)## and ##\sin(k \arctan x)##?
     
  12. Dec 18, 2013 #11

    D H

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    Then perhaps you ought to try a different approach.

    An alternative is to break the problem down into manageable parts. For example, use the fact that ##(1-2i)^{10} = (1-2i)^8(1-2i)^2##. Then use ##(1-2i)^8 = ((1-2i)^4)^2##, and then break that down even further. Eventually you'll have broken the problem down into small parts, each of which you can solve. Then build back up to where you finally find ##(1-2i)^{10}##.
     
  13. Dec 19, 2013 #12
    Let me wade through these and find where you said that . . .

    I use to have a problem with my back-hand so I took it upon myself for a while to hit the ball mostly with my backhand. Soon I didn't have a problem with my backhand.

    And that how it is in life in general: work on your problems, muscle-through them, the embarrassment, frustration, failures along the way, and soon, you too will be slammin' balls. :)
     
  14. Dec 19, 2013 #13
    I hear ya, and good advice. I actually bought a trig book to review all this stuff and during a move I lost it!!! Will have to hit amazon again lol
     
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