Finding the nth roots of a complex number

[Brian_D]

Homework Statement

Solve $\left(-2+2 \,\mathrm{I}\right)^{\frac{1}{3}}$

Relevant Equations

$$a+\mathit{bi}=r\! \left(\cos\! \left(\theta \right)+\mathrm{I} \sin\! \left(\theta \right)\right)$$
$$r=\sqrt{a^{2}+b^{2}}$$
$$\theta =\arctan\! \left(\frac{b}{a}\right)$$
$$\mathit{nth} \mathit{root} \mathit{of} a \mathit{complex} \mathit{number}=r^{\frac{1}{n}} \left(\cos\! \left(\alpha \right)+\mathrm{I} \sin\! \left(\alpha \right)\right)$$
$$\alpha=\frac{\theta+k\cdot \left(2\pi\right)}{n}$$
k=0,1,..n-1

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply $\left(r\right)^{\left(\frac{1}{n}\right)}$. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this problem, $\left(r\right)^{\left(\frac{1}{n}\right)}$ is $\left(-2+2 \,\mathrm{I}\right)^{\frac{1}{3}}$ which is 1+i, which we can see is correct because $\left(1+\mathrm{i}\right)^{3}$ equals $-2+2i$ So if 1+i is the first root, when you apply the formula you get three more supposed roots when k=0, 1, and 2. Can anyone explain this contradiction?

Secondly, when I apply the formula, it doesn't work. For example, let's see what happens when k=1. First, from the given problem, a=-2 and b=2, so theta equals arctan(-1), which is -pi/4 radians. Then the formula says to add k*2pi and divide by n, which gives us -pi/4+2pi divided by 3, which is about 1.83. So alpha is 1.83 and $r^{\frac{1}{n}} \left(\cos\! \left(\alpha \right)+\mathrm{I} \sin\! \left(\alpha \right)\right)$ equals (1+i)*(-.259 + .966i). Multiplying this out, we get $\left(-1.225+ 0.707 i\right)$, which according to the formula should be one of the cube roots. But $\left( 1.225+ 0.707 i\right)^{3}$ does not equal -2+2i, so there is something wrong. Your thoughts?

 

[Brian_D]

Homework Statement: Solve $\left(-2+2 \,\mathrm{I}\right)^{\frac{1}{3}}$
Relevant Equations: $$a+\mathit{bi}=r\! \left(\cos\! \left(\theta \right)+\mathrm{I} \sin\! \left(\theta \right)\right)$$
$$r=\sqrt{a^{2}+b^{2}}$$
$$\theta =\arctan\! \left(\frac{b}{a}\right)$$
$$\mathit{nth} \mathit{root} \mathit{of} a \mathit{complex} \mathit{number}=r^{\frac{1}{n}} \left(\cos\! \left(\alpha \right)+\mathrm{I} \sin\! \left(\alpha \right)\right)$$
$$\alpha=\frac{\theta+k\cdot \left(2\pi\right)}{n}$$
k=0,1,..n-1

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply $\left(r\right)^{\left(\frac{1}{n}\right)}$. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this problem, $\left(r\right)^{\left(\frac{1}{n}\right)}$ is $\left(-2+2 \,\mathrm{I}\right)^{\frac{1}{3}}$ which is 1+i, which we can see is correct because $\left(1+\mathrm{i}\right)^{3}$ equals $-2+2i$ So if 1+i is the first root, when you apply the formula you get three more supposed roots when k=0, 1, and 2. Can anyone explain this contradiction?

Secondly, when I apply the formula, it doesn't work. For example, let's see what happens when k=1. First, from the given problem, a=-2 and b=2, so theta equals arctan(-1), which is -pi/4 radians. Then the formula says to add k*2pi and divide by n, which gives us -pi/4+2pi divided by 3, which is about 1.83. So alpha is 1.83 and $r^{\frac{1}{n}} \left(\cos\! \left(\alpha \right)+\mathrm{I} \sin\! \left(\alpha \right)\right)$ equals (1+i)*(-.259 + .966i). Multiplying this out, we get $\left(-1.225+ 0.707 i\right)$, which according to the formula should be one of the cube roots. But $\left( 1.225+ 0.707 i\right)^{3}$ does not equal -2+2i, so there is something wrong. Your thoughts?

I don't know why my LaTeX code is not displaying, so I'll try posting it here:
$$a+\mathit{bi}=r\! \left(\cos\! \left(\theta \right)+\mathrm{I} \sin\! \left(\theta \right)\right)$$
$$r=\sqrt{a^{2}+b^{2}}$$
$$\theta =\arctan\! \left(\frac{b}{a}\right)$$
$$\mathit{nth} \mathit{root} \mathit{of} a \mathit{complex} \mathit{number}=r^{\frac{1}{n}} \left(\cos\! \left(\alpha \right)+\mathrm{I} \sin\! \left(\alpha \right)\right)$$
$$\alpha=\frac{\theta+k\cdot \left(2\pi\right)}{n}$$
k=0,1,..n-1

 

[Brian_D]

Here is my LaTeX code again
$$a+\mathit{bi}=r\! \left(\cos\! \left(\theta \right)+\mathrm{I} \sin\! \left(\theta \right)\right)$$

$$r=\sqrt{a^{2}+b^{2}}$$

$$\theta =\arctan\! \left(\frac{b}{a}\right)$$

$$\mathit{nth} \mathit{root} \mathit{of} a \mathit{complex} \mathit{number}=r^{\frac{1}{n}} \left(\cos\! \left(\alpha \right)+\mathrm{I} \sin\! \left(\alpha \right)\right)$$

$$\alpha=\frac{\theta+k\cdot \left(2\pi\right)}{n}$$

k=0,1,..n-1

 

[Brian_D]

And the homework statement is: Solve $\left(-2+2 \,\mathrm{I}\right)^{\frac{1}{3}}$ Sorry for the mess.

 

[FactChecker]

Here is my LaTeX code again
$$a+\mathit{bi}=r\! \left(\cos\! \left(\theta \right)+\mathrm{I} \sin\! \left(\theta \right)\right)$$

$$r=\sqrt{a^{2}+b^{2}}$$

$$\theta =\arctan\! \left(\frac{b}{a}\right)$$

$$\mathit{nth} \mathit{root} \mathit{of} a \mathit{complex} \mathit{number}=r^{\frac{1}{n}} \left(\cos\! \left(\alpha \right)+\mathrm{I} \sin\! \left(\alpha \right)\right)$$

$$\alpha=\frac{\theta+k\cdot \left(2\pi\right)}{n}$$

k=0,1,..n-1

That is correct. There are only n values (k=0,1,...,n-1). Your assumption that there is always a real root is wrong. If there is a real root, it will be the same as one of those n values.

 

[Brian_D]

Fabulous, thank you for clearing this up, FactChecker. But in this case, there IS a real root and, at least the way I am using the formula, it is not one of the n roots generated by the formula. So you've answered my first question, but I still don't understand why I'm not getting correct solutions from the formula.

 

[Brian_D]

Correction, there is not a real root in this case, but in any case, I'm still not getting correct solutions from the formula.

 

[FactChecker]

Correction, there is not a real root in this case, but in any case, I'm still not getting correct solutions from the formula.

Right, no real root. What roots are you getting? Are you sure they aren't correct? Sorry, I see your answers in post #1.
$r^{1/3}=1+i$ is wrong. $r$ represents the modulus. It is always a positive real number. Just take the usual cube root of that number. But that is only part of the answer. Then you must multiply $r^{1/3}$ times the 3 different complex numbers with the different arguments, $\alpha$.

Spoiler: Value of r^(1/3)

You should get $r^{1/3}=\sqrt(2)$

 

[renormalize]

And the homework statement is: Solve $\left(-2+2 \,\mathrm{I}\right)^{\frac{1}{3}}$

Here's what Mathematica gives for the roots of the equation $z^3=-2+2\,i$, expressed in both rectangular and polar form. You can use these results to verify your formula:

 

[Brian_D]

Thank you, Renormalize. However, if you check Mathematica's (rectangular) solutions by cubing each one, only the first solution produces the original expression -2+2i. I tried the same problem with Maple, and, like Mathematica, two of Maple's solutions were in an oddly complicated form, but all of Maple's solutions are correct.

 

[Brian_D]

Right, no real root. What roots are you getting? Are you sure they aren't correct? Sorry, I see your answers in post #1.
$r^{1/3}=1+i$ is wrong. $r$ represents the modulus. It is always a positive real number. Just take the usual cube root of that number. But that is only part of the answer. Then you must multiply $r^{1/3}$ times the 3 different complex numbers with the different arguments, $\alpha$.

Spoiler: Value of r^(1/3)

You should get $r^{1/3}=\sqrt(2)$[/SPOIL

Right, no real root. What roots are you getting? Are you sure they aren't correct? Sorry, I see your answers in post #1.
$r^{1/3}=1+i$ is wrong. $r$ represents the modulus. It is always a positive real number. Just take the usual cube root of that number. But that is only part of the answer. Then you must multiply $r^{1/3}$ times the 3 different complex numbers with the different arguments, $\alpha$.

Spoiler: Value of r^(1/3)

You should get $r^{1/3}=\sqrt(2)$

Thank you, FactChecker, got it. So let's see how that works out. Now for k=0, $r^{\frac{1}{n}} \left(\cos\! \left(\alpha \right)+\mathrm{I} \sin\! \left(\alpha \right)\right)$ becomes 1.26*(.966-.259i), which is (1.22-.326i). According to the formula, that should be a cube root of -2+2i, but (1.22-.326i)^3 equals 1.43-1.42i, not -2+2i. So I'm still not getting a correct solution from the formula. I can always use Maple, and maybe that is the moral of the story, but it would be nice to know why the formula is not working. Also, if you're using Mathematica, you're out of luck, as I showed in my response to renormalize.

 

[Brian_D]

I posted a reply, but somehow it got lost so I'll try again. Thank you, FactChecker, got it. So let's see how that works. For k=0, the formula now gives us 1.26*(.966-.259i)=(1.22-.326i). But when we cube this result, we get 1.43-1.42i, not -2+2i. So I'm still not getting a correct solution from the formula.

 

[Brian_D]

I see now that there was a mistake in my previous post. If the modulus is supposed to be the absolute value of -2 + 2i, then that should be 2*sqrt(2), and the cube root of that is 1.41, not 1.26. With this correction, the formula gives 1.36-.365i and the cube of that is 2-2i. For some reason the signs are still wrong, but otherwise this is a correct solution.

 

[Brian_D]

I see now that there was a mistake in my previous post. If the modulus is supposed to be the absolute value of -2 + 2i, then that should be 2*sqrt(2), and the cube root of that is 1.41, not 1.26. With this correction, the formula gives 1.36-.365i and the cube of that is 2-2i. For some reason the signs are still wrong, but otherwise this is a correct solution.

The signs are wrong because -2 +2i and 2 -2i have the same absolute value. Everything is crystal clear to me now. Thank you, FactChecker!

 

[Mark44]

And the homework statement is: Solve $\left(-2+2 \,\mathrm{I}\right)^{\frac{1}{3}}$ Sorry for the mess.

The above is an expression, not an equation or inequality. Technically speaking, you can't "solve" an expression. A better formulation of the problem would be "Find the three roots of $\left(-2+2 \,\mathrm{i}\right)^{\frac{1}{3}}$"

 

[renormalize]

Thank you, Renormalize. However, if you check Mathematica's (rectangular) solutions by cubing each one, only the first solution produces the original expression -2+2i.

All of Mathematica's roots cube to $-2+2\,i$:

Did you make an arithmetic mistake?

 

[PeroK]

I see now that there was a mistake in my previous post. If the modulus is supposed to be the absolute value of -2 + 2i, then that should be 2*sqrt(2), and the cube root of that is 1.41, not 1.26. With this correction, the formula gives 1.36-.365i and the cube of that is 2-2i. For some reason the signs are still wrong, but otherwise this is a correct solution.

1.41 is an approximation for $\sqrt 2$