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...
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