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    Could anyone help me out with this? I know the end answer is 16 but I don't know how to get there.
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    (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.
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    (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
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    (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
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      (1+i)^2 = 1+2i+i^2 = 2i

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

    Thus
      (1+i)^8 = (2i)^4 = (2^4)(i^4) = 16*1 = 16
    Thankyou for expanding it out! I really appreciate it!
 
 
 
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