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**ReallyConfused****Guest**

How do I find the cube root of a+bi.I know how to find square root of a+bi,I want to know about cube root.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

Hi;

If you know how to do it for square roots you can do it for cube roots but the equations will be tougher.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**ReallyConfused****Guest**

Can you please post the equations?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

Hi;

The general solution is very large with this method:

To get the cube root of a + bi, you would need to solve

You might want to try these videos to see another way but it is tedious too.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,336

How about converting to modulus / argument form and using

??

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,336

**Modulus argument form** continued:

I've never tried this before, so I wasn't 100% certain it would work. So I thought I'd test it out.

example:

So that worked.

I also checked it with the other two non repeating values for theta. (The three roots are symmetrically placed around the origin, 2pi/3 rads apart.)

They worked too.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi!

It may be easier to see using the exponential form for complex numbers. I'll use Q for theta.

iQ 1/n 1/n iQ/n

(re ) = r e is the principal nth root. The other n-1 roots, as Bob Bundy pointed

out, are equally spaced around the circle; that is, the increment to add to Q/n is 360/n. Keep adding

this increment to each succeeding angle until adding another increment would be a "wrap around."

i100 1/4 1/4 i100/4 i25 i(25+90) i(25+180) i(25+270)

Example: (16e ) produces 16 e = 2e , 2e , 2e , 2e

where the angles are in degrees for ease of typing.

The rectangular form for complex numbers is not as nice to work with for products, quotients and

powers and roots. The exponential form is not nice with sums and differences.

iC iD i(C+D) iC iD i(C-D) iC n n inC

(ae )(be ) = abe (ae )/(be ) = (a/b)e (ae ) = a e (Roots done above)

Integral powers can be done in rectangular form but are quite difficult if n is moderately large.

So we can switch to exponential form, do the power, and switch back to rectangular form.

Products and quotients are not so bad in rectangular form as usually taught in college algebra. It

is not worth the effort to switch to exponential to do products and quotients.

Finding roots in rectangular form? I've never seen it done.

Sums and differences in rectangular form are easy. (a+bi)+(c+di) = (a+c)+(b+d)i and

(a+bi)-(c+di) = (a-c)+(b-d)i.

I've never seen sums and differences done in exponential form.

So the more powerful the operation, the more we tend to use the exponential form for the

calculations.

There are several forms that complex numbers can be written in.

iQ

r(cosQ+isinQ) or rcisQ for short; re ; (r,Q) ; a+bi ; (a,b)

Those with r and Q are just different "wrappings" for giving the r and Q. Each is useful for

interpreting in different settings. Those with a and b are different "wrappings" for giving

the a and b of the rectangular form. a+bi is more convenient for algebraic manipulations

and (a,b) is perhaps more convenient for graphing.

Taking square roots in the exponential form gives us 180 degrees for the increment. So we

can see why square roots of a positive numbers are two real numbers (Eg sqr(4)=2 or -2).

Also we can see why square root of a negative numbers are pure imaginary numbers

(Eg sqr(-4) = 2i or -2i).

i0 1/2 i0/2 i0 1/2 i(0+180) i180

sqr(4) = sqr(4e ) = 4 e = 2e = 2. Also 4 e = 2e =-2

i180 1/2 1/2 i90 i90 i270

The principal square root of -4 is (4e ) = 4 e = 2e = 2i. The other is 2e =-2i.

Have a great day!

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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