Complex numbers?

Question

Find the fifth roots of z if z=1+2i.

i'm i suppose to solve for z^5=1+2i ?

Answer 1

y = z^5 = 1 + 2i

abs (y) = sqrt (1² + 2²) = sqrt (5) = A
arg (y) = artan (2/1) = 1.107 rad = @

y = A . exp (j.@)

z = y^(1/5) = A^(1/5) . exp (j . @ / 5)

A^(-5) = 1,1746 ...

z = 1.1746 x exp (1.107 i /5) = 1.1746 . exp ( 0.2214 i )

z = 1.1746 x cos (0.2214) + 1.1746 i x sin (0.2214)

z = 1.1746 x 0.9756 + 1.1746 i x 0.2196

z = 1.1459 + 0.25797 i

test : z^5 = 0.99979 + 1.99967i =~ 1 + 2i

Answer 2

There will be five fifth roots of 1 + 2i. The magnitude of 1 + 2i is sqrt(1^2 + 2^2) = sqrt(5). The five fifth roots will all have a magnitude of sqrt(5)^(1/5) = 5^(1/10) = 1.175 approximately. To find one of the roots, we construct a complex number whose "angle" with the real axis is 1/5 that of the original complex number. 1 + 2i makes an angle of Arctan(2) (=63.4 degrees) with the real axis. So the fifth root will make an angle of 63.4/5 = 12.69 degrees. So, the 1st fifth root is (1.175)(cos(12.69)+isin(12.69) = (1.175)(0.976+0.22i) = 1.147+0.258i. The other four roots are spaced in intervals of 360/5 = 72 degrees from the first root. The general formula will be (approximately) (1.175)(cos(12.69+72n)+isin(12.69+72n)) where n is one of the numbers 0 thru 4.

Answer 3

the way to deal with roots of complex numbers is in polar form

1+2i= 5^1/2 Arctg 2 = 1.1071 rad

So z^1/5 = 5^(1/10) <1.1071/5 +2npi/5

you´ll get five different values(roots) taking n = 0,1,2,3,4

After this you can go back to binomial form using

x=rcosfi, y =r sinfi

Answer 4

Try converting them to polar coordinates. It is much easier that way.

To get the fifth root, you just get the fifth root of the distance from origin, and divide the angle by 5.

Answer 5

either (1 + i2)^(1/5) as five identical roots
(1 + ix)^5 = 1 + i2,
or 5 possibly distinct roots,
(1 + ia)(1 + ib)(1 + ic)(1 + id)(1 + ie)

too tired now, more later maybe.