Find the modulus of (1+i)^2008 + (1-i)^2008 = ?

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- Jun 5th 2011, 10:35 AMessedraHow to find this modulus?
Find the modulus of (1+i)^2008 + (1-i)^2008 = ?

- Jun 5th 2011, 11:07 AMabhishekkgp
i guess binomial theorem should do.

- Jun 5th 2011, 11:11 AMCroat
Note that $\displaystyle (1+i)^4=(1+2i-1)^2=(2i)^2=-4$ and also $\displaystyle (1+i)^4=(1-2i-1)^2=(-2i)^2=-4$

You can probably continue yourself, no? - Jun 5th 2011, 12:02 PMmr fantastic
- Jun 5th 2011, 09:43 PMProve It
- Jun 5th 2011, 10:42 PMabhishekkgp
- Jun 5th 2011, 11:04 PMProve It
Then are you going to write out 2009 terms?

- Jun 5th 2011, 11:21 PMabhishekkgp
- Jun 8th 2011, 04:57 PMTheCoffeeMachine
Prove It, I don't understand your objection to the binomial theorem:

$\displaystyle \begin{aligned} \bigg|(1+i)^{2008}+(1-i)^{2008}\bigg| & = \bigg|\sum_{k=0}^{2008}\binom{2008}{k}i^k+\sum_{k= 0}^{2008}\binom{2008}{k}(-1)^ki^k\bigg| \\& = \bigg|2\sum_{k= 0}^{\lfloor{\frac{2008}{2}}\rfloor} \binom{2008}{2k}i^{2k}\bigg| \\& = \bigg|2\sum_{k= 0}^{1004} \binom{2008}{2k}(i^2)^k\bigg| \\& = \bigg|2\sum_{k= 0}^{1004} \binom{2008}{2k}(-1)^k\bigg|\\& = 2\sum_{k= 0}^{1004} \binom{2008}{2k}(-1)^k.\end{aligned}$ - Jun 8th 2011, 08:33 PMProve It
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 $\displaystyle \displaystyle (1 + i)^{2008}$ and $\displaystyle \displaystyle (1 - i)^{2008}$. Converting back to Cartesians enables you to add them and find the modulus. - Jun 9th 2011, 04:16 AMHallsofIvy
Mr. Fantastic's suggestion is the best. Note that the arguments of both 1+ i and 1- i are $\displaystyle \pi/4$ and 2008 is divisible by 4.

- Jun 9th 2011, 04:28 AMTheCoffeeMachine