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(1+i)^8

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1. Could anyone help me out with this? I know the end answer is 16 but I don't know how to get there.
2. (Original post by keepyourapology)
Could anyone help me out with this? I know the end answer is 16 but I don't know how to get there.
Write 1+i in polar form as sqrt(2) * (cos pi/4 + i sin pi/4).
Now raise sqrt(2) to the power 8 to give 2^(8/2) = 2^4 = 16, and note that for (cos pi/4 + i sin pi/4)^8, we can use De Moivre's Theorem to write this as (cos 8pi/4 + i sin 8pi/4) = cos 2pi + i sin 2pi = 1 + i(0) = 1. Now we multiply the 16 by the 1 to give the final answer of 16.

Alternatively if you don't know about polar form and De Moivre's Theorem yet, then just expand (1+i)^8 using the Binomial Theorem and simplify using the fact that powers of i follow the cycle 1,i,-1,-i.
3. (Original post by keepyourapology)
Could anyone help me out with this? I know the end answer is 16 but I don't know how to get there.
change 1+i into mod-arg form and use De moivre's theorem
4. (Original post by keepyourapology)
Could anyone help me out with this? I know the end answer is 16 but I don't know how to get there.
We seem to have a bit of complication creeping into people's answers again tonight!

Let's try a simpler problem: what's (1+i)^2?

Once you can do this, it should be very straightforward to work out (1+i)^4 and hence (1+i)^8

5. Thus
6. (Original post by HapaxOromenon3)
Write 1+i in polar form as sqrt(2) * (cos pi/4 + i sin pi/4).
Now raise sqrt(2) to the power 8 to give 2^(8/2) = 2^4 = 16, and note that for (cos pi/4 + i sin pi/4)^8, we can use De Moivre's Theorem to write this as (cos 8pi/4 + i sin 8pi/4) = cos 2pi + i sin 2pi = 1 + i(0) = 1. Now we multiply the 16 by the 1 to give the final answer of 16.

Alternatively if you don't know about polar form and De Moivre's Theorem yet, then just expand (1+i)^8 using the Binomial Theorem and simplify using the fact that powers of i follow the cycle 1,i,-1,-i.
Thank you so much! I haven't learnt De Moivre's Theorem yet but the second bit really helped me out!

(Original post by solC)
change 1+i into mod-arg form and use De moivre's theorem
Thank you for replying! Unfortunately, I haven't learnt De Moivre's Theorem yet.

(Original post by davros)
We seem to have a bit of complication creeping into people's answers again tonight!

Let's try a simpler problem: what's (1+i)^2?

Once you can do this, it should be very straightforward to work out (1+i)^4 and hence (1+i)^8
Thank you so much! I ended up using this method.

(Original post by newblood)

Thus
Thankyou for expanding it out! I really appreciate it!

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Updated: September 12, 2016
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