What is real part in #(i-sqrt(3))^13#?

1 Answer
George C.
Apr 25, 2018

#Re((i-sqrt(3))^13) = -4096sqrt(3)#

Explanation:

Here are a couple of methods...

Method 1

Use de Moivre's theorem:

#(cos theta + i sin theta)^n = cos n theta + i sin n theta#

Note that:

#i - sqrt(3) = 2(-sqrt(3)/2+1/2i) = 2(cos((5pi)/6)+i sin((5pi)/6))#

So:

#(i-sqrt(3))^13 = 2^13(cos((5pi)/6)+i sin((5pi)/6))^13#

#color(white)((i-sqrt(3))^13) = 2^13(cos((65pi)/6)+i sin((65pi)/6))#

#color(white)((i-sqrt(3))^13) = 2^12 * 2(cos((5pi)/6)+i sin((5pi)/6))#

#color(white)((i-sqrt(3))^13) = 2^12 * (i - sqrt(3))#

#color(white)((i-sqrt(3))^13) = 4096(i - sqrt(3))#

#color(white)((i-sqrt(3))^13) = 4096i - 4096sqrt(3)#

So the real part is #-4096sqrt(3)#

Method 2

The binomial theorem tells us that:

#(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k#

where #((n),(k)) = (n!)/((n-k)! k!)#

These binomial coefficients can be observed as the #(n+1)#th row in Pascal's triangle...

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In particular:

#(a+b)^3 = a^3+3a^2b+3ab^2+b^2#

So:

#(i - sqrt(3))^3 = i^3-3i^2sqrt(3)+3i(sqrt(3))^2-(sqrt(3))^3#

#color(white)((i - sqrt(3))^3) = -i+color(red)(cancel(color(black)(3sqrt(3))))+9i-color(red)(cancel(color(black)(3sqrt(3))))#

#color(white)((i - sqrt(3))^3) = -8i#

So:

#(i-sqrt(3))^12 = ((i-sqrt(3))^3)^4 = (8i)^4 = 8^4 i^4 = 4096#

So:

#(i-sqrt(3))^13 = (i-sqrt(3))^2 (i - sqrt(3)) = 4096(i-sqrt(3)) = 4096i - 4096sqrt(3)#

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