What is the solution to the complex equation z^3-i=0?

[brcole]

Homework Statement

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

Homework Equations

z^3-i=0

The Attempt at a Solution

z = -(i^3) = -((0+i)^3)

 

[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.

 

[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?

 

[gabbagabbahey]

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?

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

 

[gabbagabbahey]

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.

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

 

[Mentallic]

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

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"

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:

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

 

[Dick]

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"

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:

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

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.

 

[Mentallic]

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.

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.

 

[Dick]

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.

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 into begin with. I don't think it simplifies it. It's much easier to find the arguments by dividing angles by three.

 

[Mentallic]

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 into begin with. I don't think it simplifies it. It's much easier to find the arguments by dividing angles by three.

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 $$(-i)^2-4(1)(-1)=-1+4=3$$

 

[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.

 

[Mentallic]

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.

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 that's the only way I know how I am unfamiliar with this 'polar form'.

Please, don't judge me too harshly.