Complex numbers: find ((1+i*sqrt(3))/(1-i))^80 in algebraic form.

Jack10

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Dec 13, 2018
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Hi I’m wanting to find ((1+i*sqrt(3))/(1-i))^80
in algebraic form. I can get the correct answer by changing
into exp form then applying the power then changing back
however the answer provided is -2^39(1-i*sqrt(3)). So I’m
looking for the way in which I come to this answer
 
Hi I’m wanting to find ((1+i*sqrt(3))/(1-i))^80
in algebraic form. I can get the correct answer by changing
into exp form then applying the power then changing back
however the answer provided is -2^39(1-i*sqrt(3)). So I’m
looking for the way in which I come to this answer
\(\displaystyle \displaystyle{\left[ \dfrac{1 + i\sqrt{3}}{1-i}\right]^{80}}\)

= \(\displaystyle \displaystyle{\left[ \dfrac{1 + i\sqrt{3}}{1-i} * \dfrac{1+i}{1+i}\right]^{80}}\)

continue....
 
Hi I’m wanting to find ((1+i*sqrt(3))/(1-i))^80
in algebraic form. I can get the correct answer by changing
into exp form then applying the power then changing back
however the answer provided is -2^39(1-i*sqrt(3)). So I’m
looking for the way in which I come to this answer

Can you show your work, so we can see why you think it didn't lead to the correct answer (or correct form)? It isn't clear what you are lacking.
 
Hi I’m wanting to find ((1+i*sqrt(3))/(1-i))^80
in algebraic form. I can get the correct answer by changing
into exp form then applying the power then changing back
however the answer provided is -2^39(1-i*sqrt(3)). So I’m
looking for the way in which I come to this answer
Although you have been asked to so work, having spent over an hour trying to get that answer I became convinced it was incorrect even though WolframAlpha agrees with the text(SEE HERE). I was reminded that as a young Asst. Prof. teaching preCal I would spend a long trying get the form of an answer in the 'back-of-the-book'.
This is messy.
\(\displaystyle \begin{align*}\dfrac{{1 + \sqrt 3 i}}{{1 - i}} &= \dfrac{{2\exp \left( {\frac{{i\pi }}{3}} \right)}}{{\sqrt 2 \exp \left( {\frac{{ - i\pi }}{4}} \right)}}\\& = \sqrt 2 \exp \left( {\frac{{7i\pi }}{{12}}} \right) \end{align*}\) so then \(\displaystyle \begin{align*}{\left( {\dfrac{{1 + \sqrt 3 i}}{{1 - i}}} \right)^{80}} &= {\left( {\sqrt 2 \exp \left( {\frac{{7i\pi }}{{12}}} \right)} \right)^{80}}\\& = {2^{40}}\exp \left( {\frac{{2i\pi }}{3}} \right) \\&= {2^{40}}\left( {\frac{{ - 1 + i\sqrt 3 }}{2}} \right) \\&= {2^{39}}\left( { - 1 + i\sqrt 3 } \right)\end{align*}\)
 
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