How to find this modulus?

• Jun 5th 2011, 11:35 AM
essedra
How to find this modulus?
Find the modulus of (1+i)^2008 + (1-i)^2008 = ?
• Jun 5th 2011, 12:07 PM
abhishekkgp
i guess binomial theorem should do.
• Jun 5th 2011, 12:11 PM
Croat
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?
• Jun 5th 2011, 01:02 PM
mr fantastic
Quote:

Originally Posted by essedra
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.
• Jun 5th 2011, 10:43 PM
Prove It
Quote:

Originally Posted by abhishekkgp
i guess binomial theorem should do.

Are you going to write out 4018 terms?
• Jun 5th 2011, 11:42 PM
abhishekkgp
Quote:

Originally Posted by Prove It
Are you going to write out 4018 terms?

half of the terms would cancel... then using integration techniques the remaining series could be solved.
• Jun 6th 2011, 12:04 AM
Prove It
Then are you going to write out 2009 terms?
• Jun 6th 2011, 12:21 AM
abhishekkgp
Quote:

Originally Posted by Prove It
Then are you going to write out 2009 terms?

i did that but now i see its doing no good. sorry for all that.
• Jun 8th 2011, 05:57 PM
TheCoffeeMachine
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}
• Jun 8th 2011, 09:33 PM
Prove It
Quote:

Originally Posted by TheCoffeeMachine
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.
• Jun 9th 2011, 05:16 AM
HallsofIvy
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.
• Jun 9th 2011, 05:28 AM
TheCoffeeMachine
Quote:

Originally Posted by Prove 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 $\displaystyle (1 + i)^{2008}$ and $\displaystyle (1 - i)^{2008}$. Converting back to Cartesians enables you to add them and find the modulus.

Fair enough.