Finding Three Cube Roots of -1+i: Exploring Complex Roots in Math

[James889]

Hai,

I wanted to find the three cube roots of -1+i

And since the questions says `three` n = 3

So it should be something like this $$\sqrt{2}cos(\frac{3\pi}{4} + i sin(\frac{3\pi}{4})$$

But the key says that the answer is on the form $$2^{1/6}$$ why? clearly n was 3?

 

[jambaugh]

Hai,

I wanted to find the three cube roots of -1+i

And since the questions says `three` n = 3

So it should be something like this $$\sqrt{2}cos(\frac{3\pi}{4} + i sin(\frac{3\pi}{4})$$

But the key says that the answer is on the form $$2^{1/6}$$ why? clearly n was 3?

How did you go about finding the roots? Remember that when you write the radicand in polar form you must take the cube root of both the modulus an the unit factor:
$$|-1 + i | = \sqrt{1+1} = 2^{1/2}$$
So $$-1+i = 2^{1/2}u = ( 2^{1/6} v )^3$$
where $$v^3 = u$$
v and u unit complex numbers.

 

[Elucidus]

The complex number (-1 + i) is as you state $$\sqrt{2} \operatorname{cis} (3 \pi / 4)$$ where $\operatorname{cis} \theta = \cos \theta + i \sin \theta$.

A corollary to DeMoivre's Theorem states that the n, n^th-roots of a complex number $r \cdot \operatorname{cis} \theta$ are

$$\sqrt[n]{r} \operatorname{cis} \left( \frac{\theta + 2 \pi k}{n} \right) \text{ for }k=0,1, \dots ,{n-1}$$​

In your case the modulus is $\sqrt{2}$. When you take the cube root you get

$$\sqrt[3]{\sqrt{2}} = \sqrt[6]{2}$$​

I hope this helps.

--Elucidus

 

[James889]

so the other roots must be

$$2^{1/6}cos(\frac{\pi}{4}+\frac{2\pi}{6}) +i sin(\frac{\pi}{4}+\frac{2\pi}{6}) = \frac{7\pi}{12}$$

and
$$\frac{\pi}{4}+\frac{4\pi}{6} = \frac{11\pi}{12}$$

$$\frac{\pi}{4}+\frac{6\pi}{6} = \frac{15\pi}{12}$$

 

[jambaugh]

so the other roots must be

$$2^{1/6}cos(\frac{\pi}{4}+\frac{2\pi}{6}) +i sin(\frac{\pi}{4}+\frac{2\pi}{6}) = \frac{7\pi}{12}$$

and
$$\frac{\pi}{4}+\frac{4\pi}{6} = \frac{11\pi}{12}$$

$$\frac{\pi}{4}+\frac{6\pi}{6} = \frac{15\pi}{12}$$

Be careful with your ='s. I don't understand what you mean by a + i b = r when b is not zero.

Also since you are looking for cube roots shouldn't you be adding multiples of 2 pi over 3?

[edit] Also I think you got it just wrong. You should be using:
$$( \cos(t) + i \cdot \sin(t) )^n = \cos(nt) + i\cdot\sin(nt)$$
Show your work so far instead of just these bits so I don't have to work it from scratch trying to guess your method. See https://www.physicsforums.com/showthread.php?t=5374" about submitting homework!

[edit2] I tell my math students "the quickest way to get there is to take your time!" You might even have found your own error if you'd more carefully presented your work. Your "quick question" is not so quick to be answered if you don't show the details of your work.

 

[James889]

Alright,

So we have that the angle of -1 + i is $$\theta = tan^{-1}(-1) = \frac{\pi}{4}$$

The remaining roots should just be multiples by $$\frac{2\pi*k}{6}~\text{where k = 0,1,2}$$

And therefore $$r_{2}=~\frac{pi}{4} + \frac{2\pi *1}{6} \longrightarrow\frac{3\pi+4\pi}{12}$$


$$r_{3}=~\frac{pi}{4} + \frac{2\pi * 2}{6} \longrightarrow \frac{3\pi + 8\pi}{12}$$

 

[Elucidus]

Alright,

So we have that the angle of -1 + i is $$\theta = tan^{-1}(-1) = \frac{\pi}{4}$$

It is true that the tangent of the argument of (-1 + i) is -1 but the arctangent of -1 is not [itex]\pi /4[/tex]; it is $- \pi / 4$. Remember that the tangent is negative in the 2^nd and 4^th quadrants but the arctangent function can only return values in quadrant 1 and in 4 (as a negative angle).

Since the real component is negative and the imaginary component is positve, the original point lies in the 2^nd complex quadrant and thus its argument must be $3\pi /4$ as you had stated in your first post.

The rest should follow from there.

--Elucidus

 

[James889]

So the principal root is then $$2^{1/6}cos\bigg(\frac{3\pi}{4}\bigg) + isin\bigg(\frac{3\pi}{4}\bigg)$$?

 

[jambaugh]

Alright,

So we have that the angle of -1 + i is $$\theta = tan^{-1}(-1) = \frac{\pi}{4}$$

The remaining roots should just be multiples by $$\frac{2\pi*k}{6}~\text{where k = 0,1,2}$$

And therefore $$r_{2}=~\frac{pi}{4} + \frac{2\pi *1}{6} \longrightarrow\frac{3\pi+4\pi}{12}$$


$$r_{3}=~\frac{pi}{4} + \frac{2\pi * 2}{6} \longrightarrow \frac{3\pi + 8\pi}{12}$$

You need to start by writing:

$$-1 + i = r[\cos(\phi) +i\cdot \sin(\phi)]$$
then double check it is correct for your angle.

You then use the fact that you can add $$\phi + 2\pi n$$ for any integer n and you still have your original complex number. (here sufficient for n = 0, 1, 2.)

Then show how you apply the power of complex numbers in polar form. What formula or rule are you using? I do this with complex exponentials. Some texts use a CIS function which comes to the same thing some may even work with matrices which is a bit esoteric but very "correct" algebraically.

Justify your statement "The remaining roots should just be multiples by ..."

I'm not going to confirm or hint at any more answers you ask until I see some detailed work!

In future use the template for homework questions:

Homework Statement

Homework Equations

The Attempt at a Solution

In fact you should re-submit this one in this format.
----------
I'm not just being difficult for difficulty's sake. The point of the assignment is for you to go through the process carefully. Only when you understand it well then you can "be quick" and indeed you'll be both quick and correct. But I'm systematic even now on this sort of problem if its been a couple of days since I played with them because I store in my head the process I'm trying to get you to expound upon and not a memorized formula. That is the part which will transfer to other applications and make complex numbers a tool instead of a toy.

You'll get this problem shortly and kick yourself for the error. But you need to work on the more fundamental error in your method. Work through the problem as if you were explaining it to someone else. When you get an incorrect result you should start from scratch and view your steps skeptically. You need to be able to catch the errors from the front (as you work) and not from the back (after you compare your answer to a solution manual or see its not one of a multiple choice answer.)

Again remember: the quickest way to get there (= to the right answer and mastering the right method) is to take your time.