Question 1180532
Find the 3 cube roots of -8 in polar form.
<pre>
-8 = -8+0i

Graph the vector whose magnitude (modulus) is r=8, whose tail is at (0,0),
and whose tip is at (-8,0), and whose argument (angle) is &theta;=180<sup>o</sup>. 

{{{drawing(200,200,-10,10,-10,10,
line(-11,0,11,0), line(0,-11,0,11), line(-8,.01,0,.01),line(-8,-.01,0,-.01),
line(-8,-.02,0,-.02), line(-8,-.03,0,-.03),line(-8,-.04,0,-.04),
line(-8,-.1,0,-.1),line(-7.5,.6,-8,0),line(-7.5,-.5,-8,0),
red(arc(0,0,4,-4,0,180)), locate(-6,2,r=8), red(locate(-6,5.3,theta=180^o)) 

 )}}}

{{{-8}}}{{{""=""}}}{{{-8+0*i}}}{{{""=""}}}{{{r(cos(theta)+i*sin(theta))}}}{{{8(cos(180^o)^""+i*sin(180^o))}}}

Since the cube root is the 1/3 power:

{{{matrix(2,1,"",(-8)^(1/3))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",8^(1/3)(cos(180^o+360^o*n)^""+i*sin(180^o+360^o*n))^(1/3))}}}

We raise everything to the 1/3 power.  In doing so we will use deMoivre's
theorem, where we raise the magnitude (modulus 8) to the 1/3 power (i.e.,
take its cube root 2), and multiply its argument (angle) by 1/3.

{{{matrix(2,1,"",2(cos(60^o+120^o*n)^""+i*sin(60^o+120^o*n)))}}}

Now, since there are 3 cube roots, we take three consecutive integers for n.

Let n=0

{{{2(cos(60^o+120^o*0)^""+i*sin(60^o+120^o*0))}}}{{{""=""}}}{{{2(cos(60^o)+i*sin(60^o)^"")}}}

Let n=1

{{{2(cos(60^o+120^o*1)^""+i*sin(60^o+120^o*1))}}}{{{""=""}}}{{{2(cos(180^o)+i*sin(180^o)^"")}}}

Let n=2

{{{2(cos(60^o+120^o*2)^""+i*sin(60^o+120^o*2))}}}{{{""=""}}}{{{2(cos(60^o+240^o)+i*sin(60^o+240^o)^"")}}}{{{""=""}}}{{{2(cos(300^o)+i*sin(300^o)^"")}}}.

[Notice that the second one would turn out to be 2(-1+0i) or just -2, which
is the real cube root of -8.]

Edwin</pre>