Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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noelevans
2012-10-27 13:51:16

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!

bob bundy
2012-10-27 05:08:45

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

bob bundy
2012-10-26 21:20:01

How about converting to modulus / argument form and using

??

Bob

bobbym
2012-10-26 20:59:39

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.

ReallyConfused
2012-10-26 20:41:11

Can you please post the equations?

bobbym
2012-10-26 20:32:38

Hi;

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

ReallyConfused
2012-10-26 19:52:29

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.