Komplexe Gleichung lösen Kontrolle

Question

Aufgabe: z^4 -6z^2 +18=0

mein Versuch:

⇔  (z^2-3)^2   =  -9  |√

⇔   z^2  -3       =   3i |+3   ∨     z^2   -3   =  -3i  | +3

⇔   z^2            = 3+3i    ∨     z^2        = 3-3i

⇔  z^2             =  √(18)  e^^(i(arctan(1)  |√     ∨  z^2   = √(18) e^^{(i(2π-arctan(1)}   |√

⇔z_1 = 4^√(18) e^^{(i(arctan1)*(1/2))}     ∨    z_2 = 4^√(18) e^^{(i(2π-arctan1)*(1/2))}

⇔z_3 = 4^√(18) e^^{(i(2π+arctan1)*(1/2))}  ∨  z_4=  4^√(18) e^^{(i(4π-arctan1)*(1/2))}

Answer 1

Mein CAS sagt

\([x=-\sqrt{3}\,\sqrt{i+1},x=\sqrt{3}\,\sqrt{i+1},x=-\sqrt{3}\,\sqrt{1-i},x=\sqrt{3}\,\sqrt{1-i}]\)