How to find this modulus?

**essedra** · June 5th 2011, 10:35 AM

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

**abhishekkgp** · June 5th 2011, 11:07 AM

i guess binomial theorem should do.

**Croat** · June 5th 2011, 11:11 AM

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?

**mr fantastic** · June 5th 2011, 12:02 PM

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.

**Prove It** · June 5th 2011, 09:43 PM

Originally Posted by abhishekkgp

i guess binomial theorem should do.

Are you going to write out 4018 terms?

**abhishekkgp** · June 5th 2011, 10:42 PM

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.

**Prove It** · June 5th 2011, 11:04 PM

Then are you going to write out 2009 terms?

**abhishekkgp** · June 5th 2011, 11:21 PM

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.

**TheCoffeeMachine** · June 8th 2011, 04:57 PM

Prove It, I don't understand your objection to the binomial theorem:

$\begin{aligned} \bigg|(1+i)^{2008}+(1-i)^{2008}\bigg| & = \bigg|\sum_{k=0}^{2008}\binom{2008}{k}i^k+\sum_{k= 0}^{2008}\binom{2008}{k}(-1)^ki^k\bigg| \\& = \bigg|2\sum_{k= 0}^{\lfloor{\frac{2008}{2}}\rfloor} \binom{2008}{2k}i^{2k}\bigg| \\& = \bigg|2\sum_{k= 0}^{1004} \binom{2008}{2k}(i^2)^k\bigg| \\& = \bigg|2\sum_{k= 0}^{1004} \binom{2008}{2k}(-1)^k\bigg|\\& = 2\sum_{k= 0}^{1004} \binom{2008}{2k}(-1)^k.\end{aligned}$

**Prove It** · June 8th 2011, 08:33 PM

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.

**HallsofIvy** · June 9th 2011, 04:16 AM

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.

**TheCoffeeMachine** · June 9th 2011, 04:28 AM

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.

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