# Solving a Complex equation

1. Nov 24, 2008

### brcole

1. The problem statement, all variables and given/known data

Determine all z ∈ C such that z^3-i = 0.

2. Relevant equations

z^3-i=0

3. The attempt at a solution
z = -(i^3) = -((0+i)^3)

2. Nov 24, 2008

### Mentallic

Do you know how to factorize the sum/difference of 2 cubes?

$$a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$$

Once you've applied this, you will have a root and a quadratic (the 2nd factor). Just apply the quadratic formula and you will then have your 3 complex roots.

3. Nov 24, 2008

### Dick

The equation you want to solve is z^3=i, not z=(-i)^3. Write i in polar form. i=e^(pi*i/2). Does that sound familiar?

4. Nov 24, 2008

### gabbagabbahey

Technically, $$i=e^{i(\pi/2+2\pi n)}$$ where $n$ is any integer... this is why you get 3 distinct roots.

5. Nov 24, 2008

### gabbagabbahey

How exactly is this useful here?...b would be $$i^{1/3}$$, but that is exactly what he is trying to find.

6. Nov 24, 2008

### Mentallic

This reminds of how some of my classmates would say "but x^3-1 can't be factorised because we need the difference of 2 cubes" :tongue:

So we have $$z^3-i=0$$

$$i=\sqrt{-1}$$
$$i^2=-1$$
$$i^3=-\sqrt{-1}=-i$$

Thus, this can be transformed into the difference of 2 cubes such that:

$$z^3+i^3=0$$

7. Nov 24, 2008

### Dick

This is all true. You can do it that way. But you going to have a hard time applying that to the more general problem of solving z^n=C. Unless i) you can guess an nth root for C (I think that was gabbagabbahey's point) and ii) you can solve the other n-1 degree equation. I think it's better to take the general approach.

8. Nov 24, 2008

### Mentallic

Taking the general approach is all fine and dandy, but using it constantly for all questions that appear in that form... well... it can cloud your mind and you wouldn't even look twice at the question to see if there is an easier and less prone-to-failure approach.

Once again, this reminds me of my class on a similar matter. We were asked to solve for x: $$(x+4)^2=3$$ and wouldn't you know it, half the class took the 'general approach' by expanding out, simplifying, and using the quadratic equation. There were many more algebraic errors coming from the half that expanded than from the half that instantly realized the square roots could be taken from both ends.

All in all, for this question, I would recommend factorizing.

9. Nov 24, 2008

### Dick

Well, I don't. If the 'factoring' led to to a trivial solution, I would say, sure. But it doesn't. Solving the quadratic part involves finding the square root of a complex number, which puts you right back into much the same sort of a boat you were in to begin with. I don't think it simplifies it. It's much easier to find the arguments by dividing angles by three.

10. Nov 25, 2008

### Mentallic

No not quite. So we have $$z^3+i^3=0$$ (this is getting to be quite ridiculous since after all this bickering the OP is going to log back in and find the solution handed out on a platter.

factorizing: $$(z+i)(z^2-iz-1)=0$$

Now for the quadratic factor. The discriminant is $$b^2-4ac$$. This is real since $$(-i)^2-4(1)(-1)=-1+4=3$$

11. Nov 25, 2008

### Dick

You're right. My mistake. I had the i in the wrong place in the quadratic. It is a reasonable sneaky way to get a quick answer. Sorry. If the OP hasn't been back before now, they are probably not coming back.

Last edited: Nov 25, 2008
12. Nov 25, 2008

### Mentallic

I guess that disproves my point of taking an easier approach to decrease the chance of silly errors :shy:

Ahh don't be sorry. I too have a shameful secret.. Don't tell anybody, but I only took this approach to the question because thats the only way I know how I am unfamiliar with this 'polar form'.

Please, don't judge me too harshly.