SOLUTION: Find all the sixth roots of (12+5i)

Question 274605: Find all the sixth roots of (12+5i)
Found 2 solutions by CharlesG2, JBarnum:
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
Find all the sixth roots of (12+5i)
This is going to be very hard and complicated and cumbersome to figure out, for example I will show just the square root of 12+5i.
i=sqrt(-1), i^2=-1
(x+yi)^2 = 12 + 5i = a + bi, where a=12, and b=5
x^2 + 2xyi + y^2*i^2 = 12 + 5i
x^2 + 2xyi - y^2 = 12 + 5i
x^2 - y^2 + 2xyi = 12 + 5i
x^2 - y^2 = 12
2xy = 5 (this one can be turned to y=5/(2x))
x^2 - 25/(4x^2) = 12
(4x^2) * x^2 - 25 = 12 * (4x^2)
4x^4 - 25 = 48x^2
4x^4 - 48x^2 - 25 = 0
x^4 - 12x^2 - 25/4 = 0
substitute z=x^2
z^2 - 12z - 25/4 = 0
(z - 12.5)(z + 0.5) = 0
using foil: z^2 + 0.5z - 12.5z - 6.25
z = 25/2 = 12.5 or z = -1/2 = -0.5
x = +- sqrt(25)/sqrt(2) or x = +- sqrt(-1)/sqrt(2)
x = +- 5/sqrt(2) = +- 5sqrt(2)/2
or x = +- i/sqrt(2) = +- isqrt(2)/2
that is 4 possible answers for x
2xy = 5
use all 4 possible answers for x to solve for y
one:
2 * 5sqrt(2)/2 * y = 5
5sqrt(2) * y = 5
y = 5/(5sqrt(2) = 1/sqrt(2) = sqrt(2)/2
two:
2 * -5sqrt(2)/2 * y = 5
-5sqrt(2) * y = 5
-sqrt(2) * y = 1
y = -1/sqrt(2) = -sqrt(2)/2
three:
2 * isqrt(2)/2 * y = 5
isqrt(2) * y = 5
y = 5/(isqrt(2))
y = 5sqrt(2)/2i
four:
2 * -isqrt(2)/2 * y = 5
-isqrt(2) * y = 5
y = -5/(isqrt(2))
y = -5sqrt(2)/2i
going back:
(x+yi)^2 = 12 + 5i
x + yi = 5sqrt(2)/2 + (sqrt(2)/2)i
= -5sqrt(2)/2 - (sqrt(2)/2)i
= isqrt(2)/2 + (5sqrt(2)/2i)i (this can be rewritten)
--> isqrt(2)/2 + 5sqrt(2)/2
--> 5sqrt(2)/2 + (sqrt(2)/2)i
= -isqrt(2)/2 - (5sqrt(2)/2i)i (this can be rewritten)
--> -isqrt(2)/2 - 5sqrt(2)/2
--> -5sqrt(2)/2 - (sqrt(2)/2)i
now this is just the square root and it had 4 possible answers, which will take you to the 3rd root of 12+5i, since w^3 * w^3 is w^6 (w being the 12+5i)
good luck getting further if you need to...
maybe if you can solve (x+yi)^6 = 12+5i and set all the non-i terms equal to 12 and all the i terms equal to 5i ...
(x+yi)(x+yi)(x+yi)(x+yi)(x+yi)(x+yi) = 12 + 5i
(x^2 + 2xyi - y^2)(x^2 + 2xyi - y^2)(x^2 + 2xyi - y^2) = 12 + 5i
(x^4 + 2(x^3)yi - (x^2)(y^2) + 2x^3yi + 4(x^2)(y^2)(i^2) - 2x(y^3)i - (x^2)(y^2) - 2x(y^3)i - y^4)(x^2 + 2xyi - y^2) = 12 + 5i
(x^4 - (xy)^2 - 4(xy)^2 - (xy)^2 - y^4 + 2(x^3)yi + 2(x^3)yi - 2x(y^3)i - 2x(y^3)i)(x^2 + 2xyi - y^2) = 12 + 5i
(x^4 - 6(xy)^2 - y^4 + 4(x^3)yi - 4x(y^3)i)(x^2 + 2xyi - y^2) = 12 + 5i
(x^4 - 6(xy)^2 - y^4 + 4((x^3)y - x(y^3))i)(x^2 - y^2 + 2xyi) = 12 + 5i
well see this is bad enough and am still not done here...

Answer by JBarnum(2146) (Show Source): You can put this solution on YOUR website!
Just wanted to see if it was easier to read his starting solution in source code format.....lol not really but goood luck on this problem.

"maybe if you can solve and set all the non-i terms equal to 12 and all the i terms equal to 5i ...

well see this is bad enough and am still not done here..."

RELATED QUESTIONS
Find all the sixth roots of... (answered by CharlesG2)
Find all the sixth roots of... (answered by robertb)
(a) Find the sixth roots of 1. (b) Find the fifth roots of... (answered by Boreal,greenestamps)
1)how to Find the complex conjugate of (3+5i)/(1+2i) 2)how to Find all the seventh... (answered by solver91311,robertb)
the square root of 3 over 2 + 3 square roots of one sixth. show all work... (answered by jsmallt9)
two roots of a polynomial equation are 3-5i and square root of 2. a. find two additional (answered by stanbon)
The mean of 5 numbers is 11. When a sixth number is 12. Find the sixth number. (answered by robertb)
the question states given -5i is a root, determine all other roots... (answered by ReadingBoosters)
Find all the roots of... (answered by ikleyn,josgarithmetic)