How can I find the rules for deriving a complex number without a complex root?

[bobie]

Homework Statement

Wolfram gives a step to step solution for
$$\sqrt[3]{-9+46i}$$
$$→ 27+ 54i -36 - 8i → 27+ 54i -36i^2 + 8i^3 = (3+2i)^3$$
Can you tell me (where to find) the rules of this derivation?
Thanks

 

[kontejnjer]

Uhm, do you actually need to factorize the given expression under the third root, or just find what the value of the root is? If it's the latter, then it's a lot easier to convert the expression under the root into trigonometric form, and use De Moivre's theorem to find the roots (note that any n-th root of a complex number has exactly n different values, so in this case you'll get 3 different roots).

 

[haruspex]

Suppose (a+ib)³ = c+id, all integers, and you wish to deduce a and b from c and d.
Multiply that out. What do you notice about how a and b relate to factors of c and d?

 

[NascentOxygen]

Homework Statement

Wolfram gives a step to step solution for
$$\sqrt[3]{-9+46i}$$
$$→ 27+ 54i -36 - 8i → 27+ 54i -36i^2 + 8i^3 = (3+2i)^3$$
Can you tell me (where to find) the rules of this derivation?
Thanks

It's tantalizingly cryptic, isn't it? But I don't know either. The only heuristic I've come across is the original method of Bombelli himself. It's illustrated in the concluding half of the section subtitled Bombelli's Solution 5 here: http://mathfaculty.fullerton.edu/mathews/c2003/ComplexNumberOrigin.html

Possibly wolframalpha has a more direct technique for a solution having both integer components? How do you get wolframalpha to show its intermediate steps?

 

[bobie]

How do you get wolframalpha to show its intermediate steps?

Either you subscribe at $4 per month, or you can get 2 free-accesses per day opening an account.
Thanks for the nice link

 

[willem2]

If you think there is an easy solutiion:

If you compute the norm for (-9+46i)= 9^2+46^2 you get the third power of 13, so the norm of the cube root is 13. It's easy to see that there is only one way to get a sum of two squares of 13, so the cube root must be $\pm$3 $\pm$2i or $\pm$2 $\pm$ 3i.

If you sketch -9+46i in the complex plane, it's easy to see that the cube root must be in the first quadrant.

 

[bobie]

. It's illustrated in the concluding half of the section subtitled Bombelli's Solution 5 here: http://mathfaculty.fullerton.edu/mathews/c2003/ComplexNumberOrigin.html

In that very interesting article you can read that Wessel (figure 1.6):

... saw that the angular displacement of the product of two vectors should differ from the angular displacementof each factor by the same amount that the angular displacement of the other factor differs from the angulardisplacement of the standard unit vector. That's quite a mouthful; let's see what it means...

How can it be? vectors were invented a century later, how did he multiply vectors? do you know the details of that operation?

 

[BruceW]

I think Wessel did some of the early work in 'vectors'. also, I think he was mostly talking about a geometrical interpretation of complex numbers, not so much about the Euclidean vectors that we think about in modern physics. I'm not totally sure though, I don't know much about the history of it, although it's pretty interesting.

 

[bobie]

the Euclidean vectors that we think about in modern physics. ...although it's pretty interesting.

Thanks, Bruce could you explain the difference with the 'Euclidean' vectors? why Euclidean if not geometric? How do you multiply 2 Euclidean vectors?
suppose we have a vector v: x = 3 and y = 1, what is v*v in the two systems?

 

[BruceW]

when I say 'euclidean vector' I mean the modern idea of a vector (but not the abstract mathematical concept of linear vector spaces). so, cross product, dot product, vector differentiation are part of the idea of Euclidean vector. And the 'Euclidean vector' is probably what most physics or engineer undergraduates think of as simply a 'vector'.

From the very small amount of history I've read, it looks like first was the concept of complex numbers and quaternions (before Euclidean vectors), and then Euclidean vectors developed from these concepts. They all involve some idea of direction, and geometry. But they are slightly different. I think Wessel contributed to the geometric interpretation of the complex numbers, which later others developed further to get to what we know today as the Euclidean vector.

 

[bobie]

Where can I find how to multiply two such vectors? I tried "vector space" at wiki but did not find that

 

[BruceW]

http://en.wikipedia.org/wiki/Euclidean_vector gives the 'usual' concept of a vector which is most commonly known.

http://en.wikipedia.org/wiki/Vector_space gives the abstract (and more general) concept of a linear vector space, which is maybe less widely known.

edit: and complex numbers you know, the magnitudes multiply and the arguments add up.

edit: I should probably mention, the general linear vector space does not necessarily have 'multiplication' (i.e. inner product), this is an additional structure. http://en.wikipedia.org/wiki/Inner_product_space

 

[bobie]

Thanks, Bruce, but in both links there is only scalar ,dot and cross product, no trace of product of vectors the way Wessel intended for complex numbers where the length of the vector is squared and angle is doubled.

 

[BruceW]

yeah, none of these links are related to Wessel's work. Sorry, I thought you were looking for links for definitions of Euclidean vectors and linear vector spaces. Uh, for Wessel's work, I'm not sure, I can't find much easily. It looks like his ideas of adding complex numbers was similar to our modern way. I don't know about multiplication.

 

[bobie]

, I thought you were looking for links for definitions of Euclidean vectors and linear vector spaces. .

I'd like to know if you can multiply 2 Euclidean vectors (or any vectors), say v_a (3,1) * v_b (8,6) if you can, what is the result?

 

[NascentOxygen]

Wolfram gives a step to step solution for
$$\sqrt[3]{-9+46i}$$
$$→ 27+ 54i -36 - 8i → 27+ 54i -36i^2 + 8i^3 = (3+2i)^3$$
Can you tell me (where to find) the rules of this derivation?
Thanks

It may be possible to deduce the steps. Could you post WA's intermediate steps for a handful of other examples, e.g., cube root of (65+142i), (236+115i), (286-259i)

 

[bobie]

Could you post WA's intermediate steps for a handful of other examples, e.g., cube root of (65+142i), (236+115i), (286-259i)

(65+142i)= 125+150i - 60 - 8i = 125 + 150i + 60i^2 + 8i^3

(236+115i) =241/2 + 615√3/8 + 75i/8 + 123i√3/2 + 231/2 -615√3/8+ 845i/8 -123i√3/2
= ( 964 + 615√3 + 75i/8 + 492i√3 + 924 - 615√3+ 845i - 492i√3 )/8
( 964 + 615√3 + 75i/8 + 492i√3 - 924i^2 + 615√3i^2 - 845i^3 + 492i^3√3 )/8
(4+5√3)^3 + 3(4+5√3)^2 i (4√3-5) +3(4+5√3) ( i (4√3-5))^2) + (i (4√3-5))^3) /8
and that is the cube of (4+5√3 + i (4√3-5))/2
Good luck!

 

[NascentOxygen]

(65+142i)= 125+150i - 60 - 8i = 125 + 150i + 60i^2 + 8i^3

(236+115i) ... that is the cube of (4+5√3 + i (4√3-5))/2

Good luck!

I liked your first comment better, "Not easy".

EDIT: For (236+115i) I thought Wolframalpha would find this cube root (-4+5i), but instead it came up with this one (4+5√3 + i (4√3-5))/2 !

 

[Mentallic]

EDIT: For (236+115i) I thought Wolframalpha would find this cube root (-4+5i), but instead it came up with this one (4+5√3 + i (4√3-5))/2 !

The second solution has a smaller argument in the complex plane and WA seems to give a higher priority to those.

Similarly, you'll find $\sqrt[3]{-1}$ turns out to be $e^{i\pi/3}$ as opposed to $-1=e^{i\pi}$

 

[ehild]

A complex number has three cubic roots. Now you want that cubic root of Z=65+i142 which has integer components. Z=(a+bi)³.

Writhe the equation for the square of the magnitudes: |Z|²=20164=(a²+b²)³.

Taking the cubic root of both sides : 29 = a²+b².

Find a pair of numbers so the sum of squares add up to 29.

These give the possible magnitudes of a,b. Check which choice with what signs is appropriate.

ehild

 

[bobie]

I liked your first comment better, "Not easy".
!

How's it coming, did you give up, NascentOxygen?
How did they find the solution before the advent of calculators if they couldn't factorize?

 

[Mark44]

Homework Statement

Wolfram gives a step to step solution for
$$\sqrt[3]{-9+46i}$$
$$→ 27+ 54i -36 - 8i → 27+ 54i -36i^2 + 8i^3 = (3+2i)^3$$
Can you tell me (where to find) the rules of this derivation?
Thanks

There's a sign error that I don't think anyone noticed. 27+ 54i -36 - 8i = 27+ 54i + 36i^2 + 8i^3.

 

[bobie]

= 27+ 54i + 36i^2 + 8i^3.

Thanks for the correction, Mark.
If I cannot factorize, suppose I have
$x^3- 3*5 x = -22 , (\sqrt{11^2 - 5^3} = 2i)$

can I find the solution of $\sqrt[3]{-11-2i}$ without a computer or a calculator, just with logarithm or other tables?
is it 2i or -2i?, what equation generates -11+2i?
Can someone show me how to proceed?
And can I find the result with a pocket calculator that has the ln- e^ function?

Thanks

 

[NascentOxygen]

How's it coming, did you give up, NascentOxygen?

I never give up. This is merely a pause while awaiting inspiration.

 

[NascentOxygen]

can I find the solution of $\sqrt[3]{-11-2i}$ without a computer or a calculator, just with logarithm or other tables?

You are interested in the root having both integer components? The technique earlier illustrated by ehild works well. Thanks ehild!

√(121+4)= (√ (a² + b²))³

→ 5 = a² + b²

and there aren't many candidates for a & b here

 

[bobie]

You are interested in the root having both integer components?

I would like to learn how to proceed in general.I am just learning,
as far as I know , if p/3 ^3 > q/2 ^2 the root is to be found by e^(q/2 *i √(p/3^3-q/2^2))
Is this correct?
How do I decide the sign of the imaginary part? ehild says: choose the appropriate sign, so , isn't there a rule? why isn't the i sign always negative?
How do I proceed with pencil and paper and log/trig tables?
How do I proceed with a calculator?

Thanks, N.O. don't give up!

 

[ehild]

A complex number z=a+ib has three cubic roots, and you find them in general case with the following procedure:

Write z in the Euler form with magnitude and angle: z=|z|e^iθ. |z|=sqrt(a²+b²), cosθ=a/|z|, sinθ=b/|z|, and tanθ=b/A.

The cubic root of z is |z|^1/3 e^{iθ/3±k*2π/3)}=|z|^1/3(cos(θ/3±k*2π/3)+isin(θ/3±k*2π/3)). (k=0, 1, 2 or 0,-1,1)

If you prefer the angles in degrees, the cubic root is |z|^1/3(cos(θ/3±k*120)+isin(θ/3±k*120)). (k=0, 1, 2 or 0,-1,1)

Calculators did not exist when I was a student. We used tables and slide rules [PLAIN]http://en.wikipedia.org/wiki/Slide_rule, and also did calculation with pencil and paper.

For example, you want the cubic roots of -11-2i.

|z|= √(121+4)=√125. You will need the cubic root of the square root of 125, that is the 6th root, 125^1/6 For that use that log (125^1/6 ) = 1/6 log (125). Find log 125 in the tables, divide by 6 and find the number whose logarithm is equal to that.
In this case 125=5³, so its 6th root is equal to √5=2.236..

tanθ=2/11 you do the division by hand, it is 0.181818..

Find the corresponding angle in the trigonometric tables, (you get it usually in degrees), between -90°and 90°. It is 10.305°, (or you get it in degrees and minutes) but the angle is in the third quadrant as both the sine and the cosine are negative, so you add 180°: θ=190.305°.
Take one third: θ/3=63.435°.
With all of these, one root is 2.236(cos(63.435°)+isin(63.435°))=1.00+2.00i

The next root is 2.236(cos(183.435°)+isin(183.435°)) The angle is in the third quadrant. Subtract 180°, find the cosine and sine in the tables, and take both of them with negative sign.

2.236(cos(183.435°)+isin(183.435°))=2.236(-cos(3.435°)-isin(3.435°))= -2.232 - 0.134 i

You get the third root with angle 303.435°. It is in the fourth quadrant, the sine is negative, the cosine is positive. Find the cosine and sine of 56.565. The third root is 1.232 - 1.866 i.

Calculate the third power of all of them, it has to be close to -11-2i.


ehild

 

[bobie]

Thanks, ehild, for th excellent explanation,
but I do not how to decide the sign: if

$x^3- 3*5 x = -22 , → (\sqrt{-11^2 - 5^3} = \sqrt{-4})$
should I take the cubic root of -11+ or - 2i?


Thanks

 

[ehild]

You meant to find the roots of equation x³-15x=-22, as x=s-t, where

3st=-15 and s³-t³=-22.

t⁶-22t³+125=0, a quadratic in t³.

So $t^3=11\pm \sqrt{(-11)^2-125}=\sqrt{-4}=11\pm 2i$. The two values are complex conjugates, and their cubic roots are also complex conjugate pairs. .

You find that a cubic root of 11+2i is t1= -1-2i.

A cubic root of 11-2i is t2=-1+2i = t1^*.

You can choose any of them, they will yield x values, complex conjugate to each other. If an equation has all real coefficients, the complex roots appear in pairs - one is complex conjugate of the other. You can choose either of the t-s, you will get a root of the original equation.

See: s1=-5/t1=-5/(-1-2i)=1-2i. x1=s1-t1=1-2i+1+2i=2. s2=-5/t2=-5/(-1+2i)=1+2i, x2=s2-t2=2.

ehild

 

[bobie]

Thanks for your invaluable help, ehild,
Can you tell me how to find the third (imaginary) root when the discriminant = 0?
Wolframalpha only gives two http://www.wolframalpha.com/input/?i=x^3-27x=54

 

[ehild]

There is no complex root. Two roots coincide, x1=x2=-3, x3=6.

ehild