SOLUTION: What complex number equals (1 + isqrt{3})^6?

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Question 1085385: What complex number equals (1 + isqrt{3})^6?
Found 2 solutions by ikleyn, natolino_2017:
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
The complex number 1 + i*sqrt(3) has the modulus r = sqrt%281%5E2+%2B+%28sqrt%283%29%29%5E2%29 = sqrt%281%2B3%29 = sqrt%284%29 = 2

and the argument alpha = arctan%28sqrt%283%29%29 = pi%2F3.



Therefore, according to de Moivre formula,

%281+%2B+i%2Asqrt%283%29%29%5E6 = r%5E6%2A%28cos%286%2Aalpha%29+%2B+i%2Asin%286%2Aalpha%29%29 = 2%5E6%2A%28cos%282pi%29+%2B+i%2Asin%282pi%29%29 = 2%5E6 = 64.

Answer. 64.


There is a bunch of lessons on complex numbers in this site
    - Complex numbers and arithmetical operations on them
    - Complex plane
    - Addition and subtraction of complex numbers in complex plane
    - Multiplication and division of complex numbers in complex plane
    - Raising a complex number to an integer power
    - How to take a root of a complex number
    - Solution of the quadratic equation with real coefficients on complex domain
    - How to take a square root of a complex number
    - Solution of the quadratic equation with complex coefficients on complex domain

    - Solved problems on taking roots of complex numbers
    - Solved problems on arithmetic operations on complex numbers
    - Solved problem on taking square root of complex number
    - Miscellaneous problems on complex numbers
    - Advanced problem on complex numbers
    - A curious example of an equation in complex numbers which HAS NO a solution


Also, you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Complex numbers".



Answer by natolino_2017(77) About Me  (Show Source):
You can put this solution on YOUR website!
we have two ways (at least) to answer this:
i) using Newton binomial:
(1+isqrt(3))^6 = (6C0)(isqrt(3))^0 + (6C1)(isqrt(3))^1 +(6C2)(isqrt(3))^2 + (6C3)(isqrt(3))^3 +(6C4)(isqrt(3))^4 + (6C5)(isqrt(3))^5 + (6C6)(isqrt(3))^6
= 1 + 6sqrt(3)*i - 45 - 60sqrt(3)i + 135 + 54sqrt(3)i - 27 = 64
ii) Using the Moivre's formula:
(1+isqrt(3)) = 2cis(60°)
so (1+isqrt(3))^6 = 2^6(cis(6*60°) = 64cis(360) =64cis(0) =64(cos(0) +i sin(0)) = 64.
@natolino_