Complex numbers - polar form - does this work (indices) ?

[JPC]

Complex numbers - polar form - does this work (indices) ?

hey

i haven't studied in class complex numbers yet, but i know some of the basis , and i was wondering if something i saw in complex numbers was true :

polar form :
let 'a' be the angle
and x the length (dont know how to call it in english)
: (x, a)

ok so now is this true :

-> ( (x, a) )^n = (x^n, a * n)

-> sqroot( (x, a) ) = (sqroot(x), a/2) or (sqroot(x), pi + a/2)

 

[dextercioby]

The first formula is true. While for the second, the one following "or" is false.

 

[mathman]

The first formula is true. While for the second, the one following "or" is false.

JPC is correct. For angle a, to get nth roots, use (a+2kpi)/n for k=0,1,..,n-1.

Therefore for sqrt, need k=0,1.

 

[JPC]

ok , and now , is this true for cube root , i only checked it on xroot(3, i) yet :

polar :
xroot(3, (x, a) ) =

( xroot(3, x) , a / 3 )
or
( xroot(3, x) , pi - a/3 )
or
( xroot(3, x) , a + pi )

im especially not sure for the "( xroot(3, x) , a + pi )", because i can't really find a rule for it.
finding cube root of i :

(a+ bi)^3 = i :

a^3 - 3ab² = 0
3a²b - b^3 = 1

a ( a² - 3b²) = 0
a = 0 or a² = 3b²
a = sqroot(3) * b , or a = - sqroot(3)* b or a = 0

i replace in the second equation a² by 3b²
and find b = 1 / 2
and find a= sqroot(3) / 2

now, if a=0 , then we end up in the second equation with :
-b^3 = 1
b= -1

giving me 3 solutions :

+ sqroot(3) / 2 + 0.5i
or
- sqroot(3) / 2 + 0.5i
or
-i

 

[mathman]

polar :
xroot(3, (x, a) ) =

( xroot(3, x) , a / 3 )
or
( xroot(3, x) , pi - a/3 )
or
( xroot(3, x) , a + pi )

Your notation is very confusing:

In any case if (x,a) represents a number in polar coordinates, the three cube roots are
(y,a/3), (y,(a+2pi)/3),(y,(a+4pi)/3), where y is the cube root of x.

 

[JPC]

Oh yeah, works completely with cube root of i and others, thanks

c : an Imaginary Number)

Now maybe ill try xroot(4, c), with c = i

would it end up with 4 solutions , (do sqroot of c once, then do sqroot to each of the first sqroot answers) ?
or would there only be 2 answers ? (out of the 4 answers of that way, 2 by 2 they end up the same)

 

[mathman]

Now maybe ill try xroot(4, c), with c = i

Your notation is confusing - I have never seen anything like it. Could you explain what it means? If c is supposed to be angle, it has to be a real number.

 

[JPC]

well like square root of a number c is :
xroot(2, c)

if u have xroot(a, b) means :

...___
a \/ b

this notation is from my HP calculator

 

[mathman]

It looks like xroot(a,b) means (in standard notation) b^1/a. Here b is any complex number, but the form of b has not be specified. There are two usual ways:

1) b=x+iy, where x and y are real.

2) b=r(e^iz), where r is non-negative and z is an angle between 0 and 2pi.

The question I was trying to answer previously iinvolves using the second representation.