Four roots of this equation

[adele]

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: 2^1/2 (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?

 

[galactus]

I believe the trouble may lie in your modulus.

Try writing your forms as \(\displaystyle \frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}i\) and so on.

Your modulus is \(\displaystyle \sqrt{2}\) instead of 2.

i.e \(\displaystyle \sqrt{2}\left(cos(\frac{\pi}{3})+i\cdot sin(\frac{\pi}{3})\right)\)

 

[Deleted member 4993]

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: 2^1/2 (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?

2^1/2 (cos(1/2*pi/3) + i sin(1/2*pi/3))

\(\displaystyle = \ \sqrt{2} \cdot [cos(\frac{\pi}{6}) + i \cdot sin(\frac{\pi}{6})]\)

\(\displaystyle = \ \sqrt{2} \cdot [\frac{\sqrt{3}}{2}+ i \frac{1}{2} ] \)

\(\displaystyle = \ \dfrac{\sqrt{6}}{2}+ i \dfrac{\sqrt{2} }{2} \)

 

[lookagain]

\(\displaystyle < < \)z⁴+2z²+4=0 \(\displaystyle < < \) Find the four roots.

---

So i start of normal with substitution.

\(\displaystyle > > \)(z²)²-2z²+4=0 \(\displaystyle < < \) and let u=z²

u²-2u+4=0

adele,

which equation is it supposed to be?

You have two different equations here. The signs of
the coefficients of the middle terms are opposite
from each other.

 

[adele]

adele,

which equation is it supposed to be?

You have two different equations here. The signs of
the coefficients of the middle terms are opposite
from each other.

Yup i just noticed, fixed it. I just "miss-typed" (+) in the first one. Shud be (-) there like the other one, its same equation.

Btw :Subhotosh Khan and Galactus thanks for help, was easier if just wrote them that way and more "cleaner". Don't think i can get the roots more simplifid then that.

 

[adele]

Still abit confused, i end up with some same roots.

because these roots z=-√(1-i√3) and -√(1+i√3), (the hole thing is -√) my first tought was just switch the operator in () with normal rules. but then i end up with same roots as z= √(1+i√3) and √(1-i√3). So my question is: can i just write the oppsite of those at the end? like negative the first root and second root

first root: √6/2 + i√2/2 and just do -(√6/2 + i√2/2) with third root ?

and second root:
modulus √(1² +(-√3)²) = -2?
argument: tan = -√3/1 = pi/-3?


-2^(1/2) (cos(-1/2*pi/-3) + i sin(-1/2*pi/-3))
= √-2(cos(-pi/-6)+i sin(-pi/-6)
= √-2*(√3/2+i(1/2)
= √-6/2+i*(√-2/2)


second root: √-6/2+i*(√-2/2) and just write fourth root like -(√-6/2+i*(√-2/2)?

so the four roots are: z = √6/2 + i√2/2, z= -(√6/2 + i√2/2) , z= √-6/2+i*(√-2/2) and z = -(√-6/2+i*(√-2/2) ?

those this even look right?