Complex roots
By Anonymous on Friday, May 11, 2001 - 03:14 pm :
How would you solve this equation, giving the solution in the form r eiq
The equation being z3 + 8i = 0

Thanks
By Olof Sisask (P3033) on Friday, May 11, 2001 - 05:34 pm :
Do you know about Argument, Modulus form? i.e. [r,q]?
By Anonymous on Friday, May 11, 2001 - 05:52 pm :

Umm.. I think I know a little about it. But not sure of what to do.


Can you say that z = r(cosq+ isinq)

then, z3 = r3 (cos3q+ isin3q),

since z3 = 0 + (-8i), we can equate real and imaginary? to get

1) r3 cos3q = 0

2) r3 sin3q = -8


but not sure of what to do after this?

thanks for your help Olof.

By Olof Sisask (P3033) on Friday, May 11, 2001 - 06:17 pm :
Something that's very useful when working with questions like these, is modulus argument form, i.e. [r, q], where r is the modulus and q the argument. You can show using DeMoivre's that [r, q]t = [rt,tq]. In this case we have

z3 = -8i = [8, -p/2 + 2pn] (where n is an integer - can you see why we add this 2pn to the argument?).

Therefore z = [8, -p/2 + 2pn]1/3

= [81/3 , (-p/2 + 2pn)*1/3]

= [2, (2pn)/3 - p/6]

Now set n = 0, 1, 2 and you obtain 3 different values for q, which you can plug into your r eiq.
Hope this helps,
Olof.

By Kerwin Hui (Kwkh2) on Monday, May 14, 2001 - 02:40 pm :

Another way is to spot the solution 2i, and so obtain a quadratic for the other 2 roots.

Kerwin