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 e^{i θ}

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, θ]

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 (\cos θ + i \sin θ)

then, $z^{3} = r^{3} (\cos 3 θ + i \sin 3 θ)$ ,

since $z^{3} = 0 + (- 8 i)$ , we can equate real and imaginary? to get

1) $r^{3} \cos 3 θ = 0$

2) $r^{3} \sin 3 θ = - 8$

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, θ]

, where

r

is the modulus and

θ

the argument. You can show using DeMoivre's that

[r, θ]^{t} = [r^{t}, t θ]

. In this case we have

$z^{3} = - 8 i = [8, - π / 2 + 2 π n]$ (where $n$ is an integer - can you see why we add this $2 π n$ to the argument?).

Therefore $z = [8, - π / 2 + 2 π n]^{1 / 3}$

$= [81 / 3, (- π / 2 + 2 π n) * 1 / 3]$

$= [2, (2 π n) / 3 - π / 6]$

Now set $n = 0$ , 1, 2 and you obtain 3 different values for $θ$ , which you can plug into your $r e^{i θ}$ .
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