z4-2z2+4=0 Find the four roots.
So i start of normal with substitution.
(z²)²-2z²+4=0 and let u=z²
u²-2u+4=0
u= 1+i√3 and u= 1-i√3
And i find z
z²= 1+i√3 and z²= 1-i√3
My current four roots : z= √(1+i√3), z= -√(1+i√3), z= √(1-i√3), z=-√(1-i√3)
Then i need to simplify them using DeMoivres Theorem: first i need to write them as polarform:
first root is z= √(1+i√3) write it as z= (1+i√3)1/2 .
modulus: √(12 + √32 ) = 2
argument: tan = √3/1 = pi/3
and polarform w=( 2*(cos(pi/3) + i sin(pi/3) )1/2 ,and then with the theorem i can write: 21/2 (cos(1/2*pi/3) + i sin(1/2*pi/3)) = 1.412 + 0i and here i stop, this can't be right answer, maple comes up with diffrent answer aswell.
So my question is, what have i done wrong or missed out on?
So i start of normal with substitution.
(z²)²-2z²+4=0 and let u=z²
u²-2u+4=0
u= 1+i√3 and u= 1-i√3
And i find z
z²= 1+i√3 and z²= 1-i√3
My current four roots : z= √(1+i√3), z= -√(1+i√3), z= √(1-i√3), z=-√(1-i√3)
Then i need to simplify them using DeMoivres Theorem: first i need to write them as polarform:
first root is z= √(1+i√3) write it as z= (1+i√3)1/2 .
modulus: √(12 + √32 ) = 2
argument: tan = √3/1 = pi/3
and polarform w=( 2*(cos(pi/3) + i sin(pi/3) )1/2 ,and then with the theorem i can write: 21/2 (cos(1/2*pi/3) + i sin(1/2*pi/3)) = 1.412 + 0i and here i stop, this can't be right answer, maple comes up with diffrent answer aswell.
So my question is, what have i done wrong or missed out on?
Last edited: