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

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- Jun 5th 2011, 11:35 AMessedraHow to find this modulus?
Find the modulus of (1+i)^2008 + (1-i)^2008 = ?

- Jun 5th 2011, 12:07 PMabhishekkgp
i guess binomial theorem should do.

- Jun 5th 2011, 12:11 PMCroat
Note that and also

You can probably continue yourself, no? - Jun 5th 2011, 01:02 PMmr fantastic
- Jun 5th 2011, 10:43 PMProve It
- Jun 5th 2011, 11:42 PMabhishekkgp
- Jun 6th 2011, 12:04 AMProve It
Then are you going to write out 2009 terms?

- Jun 6th 2011, 12:21 AMabhishekkgp
- Jun 8th 2011, 05:57 PMTheCoffeeMachine
Prove It, I don't understand your objection to the binomial theorem:

- Jun 8th 2011, 09:33 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 and . Converting back to Cartesians enables you to add them and find the modulus. - Jun 9th 2011, 05:16 AMHallsofIvy
Mr. Fantastic's suggestion is the best. Note that the arguments of both 1+ i and 1- i are and 2008 is divisible by 4.

- Jun 9th 2011, 05:28 AMTheCoffeeMachine