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Math Help - How to find this modulus?

  1. #1
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    How to find this modulus?

    Find the modulus of (1+i)^2008 + (1-i)^2008 = ?
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  2. #2
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    i guess binomial theorem should do.
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    Note that (1+i)^4=(1+2i-1)^2=(2i)^2=-4 and also (1+i)^4=(1-2i-1)^2=(-2i)^2=-4

    You can probably continue yourself, no?
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  4. #4
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    Quote Originally Posted by essedra View Post
    Find the modulus of (1+i)^2008 + (1-i)^2008 = ?
    1 + i = \sqrt{2} cis\left( \frac{\pi}{4}\right) and 1 - i = \sqrt{2} cis\left( \frac{-\pi}{4}\right).

    Now use deMoivre's Theorem, add and simplify.
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  5. #5
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    Quote Originally Posted by abhishekkgp View Post
    i guess binomial theorem should do.
    Are you going to write out 4018 terms?
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Are you going to write out 4018 terms?
    half of the terms would cancel... then using integration techniques the remaining series could be solved.
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  7. #7
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    Then are you going to write out 2009 terms?
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  8. #8
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Prove It View Post
    Then are you going to write out 2009 terms?
    i did that but now i see its doing no good. sorry for all that.
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    Prove It, I don't understand your objection to the binomial theorem:


    \begin{aligned} \bigg|(1+i)^{2008}+(1-i)^{2008}\bigg| & = \bigg|\sum_{k=0}^{2008}\binom{2008}{k}i^k+\sum_{k=  0}^{2008}\binom{2008}{k}(-1)^ki^k\bigg| \\& = \bigg|2\sum_{k= 0}^{\lfloor{\frac{2008}{2}}\rfloor} \binom{2008}{2k}i^{2k}\bigg| \\& = \bigg|2\sum_{k= 0}^{1004} \binom{2008}{2k}(i^2)^k\bigg| \\& = \bigg|2\sum_{k= 0}^{1004} \binom{2008}{2k}(-1)^k\bigg|\\& =  2\sum_{k= 0}^{1004} \binom{2008}{2k}(-1)^k.\end{aligned}
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  10. #10
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    Quote Originally Posted by TheCoffeeMachine View Post
    Prove It, I don't understand your objection to the binomial theorem:


    \begin{aligned} \bigg|(1+i)^{2008}+(1-i)^{2008}\bigg| & = \bigg|\sum_{k=0}^{2008}\binom{2008}{k}i^k+\sum_{k=  0}^{2008}\binom{2008}{k}(-1)^ki^k\bigg| \\& = \bigg|2\sum_{k= 0}^{\lfloor{\frac{2008}{2}}\rfloor} \binom{2008}{2k}i^{2k}\bigg| \\& = \bigg|2\sum_{k= 0}^{1004} \binom{2008}{2k}(i^2)^k\bigg| \\& = \bigg|2\sum_{k= 0}^{1004} \binom{2008}{2k}(-1)^k\bigg|\\& =  2\sum_{k= 0}^{1004} \binom{2008}{2k}(-1)^k.\end{aligned}
    My objection is that your answer is not fully simplified. To get full marks I'm sure that the OP would need to get a single number as their answer, not a sum (after all, a modulus represents the LENGTH of a complex number). And I know that you are not going to physically sum 1005 terms...

    Meanwhile, using DeMoivre's Theorem, as Mr F suggested, enables you to quickly evaluate \displaystyle (1 + i)^{2008} and \displaystyle (1 - i)^{2008}. Converting back to Cartesians enables you to add them and find the modulus.
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  11. #11
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    Mr. Fantastic's suggestion is the best. Note that the arguments of both 1+ i and 1- i are \pi/4 and 2008 is divisible by 4.
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  12. #12
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    Quote Originally Posted by Prove It View Post
    My objection is that your answer is not fully simplified. To get full marks I'm sure that the OP would need to get a single number as their answer, not a sum (after all, a modulus represents the LENGTH of a complex number). And I know that you are not going to physically sum 1005 terms...

    Meanwhile, using DeMoivre's Theorem, as Mr F suggested, enables you to quickly evaluate \displaystyle (1 + i)^{2008} and \displaystyle (1 - i)^{2008}. Converting back to Cartesians enables you to add them and find the modulus.
    Fair enough.
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