# Thread: How to find this modulus?

1. ## How to find this modulus?

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

2. i guess binomial theorem should do.

3. 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?

4. 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.

5. Originally Posted by abhishekkgp
i guess binomial theorem should do.
Are you going to write out 4018 terms?

6. 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.

7. Then are you going to write out 2009 terms?

8. 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.

9. 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}

10. 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.

11. 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.

12. 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.