Question 1170262
find the three cube roots of 6-2i and express the roots in polar form. 

First we plot the point 6-2i which is the point (6,-2), x=6, y=-2
We draw a vector from(0,0) to (6,-2).

{{{drawing(3200/7,400,-7,9,-7,7,graph(3200/7,400,-7,9,-7,7),locate(3,.71,x=6),
locate(6.1,-.7,y=-2), green(line(6,0,6,-2),line(0,0,6,0)),line(0,0,6,-2),
red(arc(0,0,2.3,-2.3,0,342)),locate(2,-1.5,r=2sqrt(10)),locate(-1.2,1.2,theta))

 )}}}

The hypotenuse is {{{r=sqrt(x^2+y^2)=sqrt(6^2+(-2)^2)=sqrt(36+4)=sqrt(40)=sqrt(4*10)=2sqrt(10)}}}, but we'll leave it {{{sqrt(40)}}} for now.

Calculate θ by {{{tan(reference)=2/6=1/3)}}}, so {{{theta=tan^(-1)(1/3)}}}


So 6-2i in polar form is 

{{{6-2i=r(cos(theta)+i*sin(theta)^"")=sqrt(40)(cos(tan^(-1)(1/3))+i*sin(tan^(-1)(1/3))^"")}}}

We can add any integer times 360° to the angle without changing the value:

{{{6-2i=sqrt(40)(cos(theta+360^o*n)+i*sin(theta+360^o*n)^"")=sqrt(40)(cos(tan^(-1)(1/3)+360^o*n+360^o*n)+i*sin(tan^(-1)(1/3)+360^o*n)^"")}}}

{{{6-2i=sqrt(40)(cos(tan^(-1)(1/3)+360^o*n+360^o*n)+i*sin(tan^(-1)(1/3)+360^o*n)^"")}}}

We raise both side to the 1/3 power, the same as taking cube root, and write
{{{sqrt(40)}}} as the 1/2 power of 40

{{{matrix(2,1,"",(6-2i)^(1/3)=(40^(1/2)(cos(tan^(-1)(1/3)+360^o*n)+i*sin(tan^(-1)(1/3)+360^o*n)^"")^"")^(1/3))}}}

Next we use DeMoivre's theorem, which is to raise the modulus to the 1/3
power and multiply the argument by 1/3:

{{{matrix(2,1,"",(6-2i)^(1/3)=(40^((1/2)(1/3))(cos(expr(1/3)tan^(-1)(1/3)+120^o*n)+i*sin(expr(1/3)tan^(-1)(1/3)+120^o*n)^"")))}}}

{{{matrix(2,1,"",(6-2i)^(1/3)=(40^(1/6)(cos(expr(1/3)tan^(-1)(1/3)+120^o*n)+i*sin(expr(1/3)tan^(-1)(1/3)+120^o*n)^"")))}}}

{{{root(6,40)(cos(expr(1/3)tan^(-1)(1/3)+120^o*n^"") - i*sin(expr(1/3)tan^(-1)(1/3)+120^o*n^"")^"")}}} where n=0,1,2

[Write all three values out].

Or if you want it in radians, substitute {{{2pi/3}}} for 120°

Edwin</pre>