SOLUTION: please help me solve this equation. Find the cube roots of z if z = 5(cos225degrees + j sin 225degress)

z = 5(cos225° + j sin225°)

Now if you add 350° any number of times, you have the
same trig functions, therefore we can write 225°+360°k for any integer k.

z = 5[cos(225°+360°k) + j·sin(225°+360°k)]

The rule is.  To take the nth root of r(cosq+i·sinq), take the nth root 
of the modulus r, and divide the argument q by n.

So,

To take the cube (3rd) root (cube root) of

z = 5[cos(225°+360°k) + j·sin(225°+360°k)]

Take the cube root of the modulus 5, and divide the argument 225°+360°k
by 3.  Dividing 225°+360°k is 75²+120²k

 __    _  
∛z = ∛5
  
There are 3 unique different cube roots of a complex number, so we
substitute 3 consecutive integeres for k and evaluate them.  We will
pick the easiest three k = 0, 1,and 2.

Let k = 0

 __    _  
∛z = ∛5

 __    _  
∛z = ∛5 = .4426 + 1.6517j

Let k = 1

 __    _  
∛z = ∛5

 __    _  
∛z = ∛5 = -1.6517 + .4426j

Let k = 2

 __    _  
∛z = ∛5

 __    _  
∛z = ∛5 = 1.2091 + 1.2091j

Edwin