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(1+i)^8

Could anyone help me out with this? I know the end answer is 16 but I don't know how to get there. :frown:
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. :frown:


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.
Reply 2
Avatar for solC
solC
14
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. :frown:


change 1+i into mod-arg form and use De moivre's theorem
Reply 3
Avatar for davros
davros
16
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. :frown:


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 :smile:
(1+i)2=1+2i+i2=2i (1+i)^2 = 1+2i+i^2 = 2i

Thus
(1+i)8=(2i)4=(24)(i4)=16∗1=16 (1+i)^8 = (2i)^4 = (2^4)(i^4) = 16*1 = 16
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! :smile:

Original post by solC
change 1+i into mod-arg form and use De moivre's theorem


Thank you for replying! :smile: Unfortunately, I haven't learnt De Moivre's Theorem yet. :frown:

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 :smile:


Thank you so much! :smile: I ended up using this method. :smile:

Original post by newblood
(1+i)2=1+2i+i2=2i (1+i)^2 = 1+2i+i^2 = 2i

Thus
(1+i)8=(2i)4=(24)(i4)=16∗1=16 (1+i)^8 = (2i)^4 = (2^4)(i^4) = 16*1 = 16


Thankyou for expanding it out! :smile: I really appreciate it! :biggrin:

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