Question 984942
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Notice that     -64 = 64(cos(180°) + i*sin(180°)).


The sixth roots of -64 are 6 complex numbers


1)  2(cos(30°) + i*sin(30°)) = {{{2(sqrt(3)/2 + i*(1/2))}}} = {{{sqrt(3) + i}}};                                                 (Notice that 30° = 180°/6)


2)  2(cos(30°+60°) + i*sin(30°+60°)) = 2(cos(90°) + i*sin(90°)) = {{{2*0 + 2*i)}}} = {{{2i}}};       (Notice that 60° = 360°/6)


3)  2(cos(30°+120°) + i*sin(30°+120°)) = 2(cos(150°) + i*sin(150°)) = {{{2*((-sqrt(3)/2) + i*(1/2))}}} = {{{-sqrt(3) + i)}}};


4)  2(cos(30°+180°) + i*sin(30°+180°)) = 2(cos(210°) + i*sin(210°)) = {{{2*((-sqrt(3)/2) + i*(-1/2))}}} = {{{-sqrt(3) - i)}}};


5)  2(cos(30°+240°) + i*sin(30°+240°)) = 2(cos(270°) + i*sin(270°)) = {{{2*(0 - i)}}} = {{{-2i)}}};


6)  2(cos(30°+300°) + i*sin(30°+300°)) = 2(cos(330°) + i*sin(330°)) = {{{2*((sqrt(3)/2) + i*(-1/2))}}} = {{{sqrt(3) - i)}}}.


If you want to see my lessons on complex numbers in this site, look in this 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/Review-of-lessons-on-complex-numbers.lesson>REVIEW of lessons on complex numbers</A>

and especially in this one 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/How-to-take-a-root-of-a-complex-number.lesson>How to take a root of a complex number</A>.