What are the solutions to the equation z^6-(3+2i)z^3+2+2i=0 ?

The solutions to the equation z^6-(3+2i)z^3+2+2i=0 are

Find the zeros of z^6-(3+2i)z^3+2+2i=0

The zeros of z^6-(3+2i)z^3+2+2i=0 are

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z^{6} - (3 + 2 i) z^{3} + 2 + 2 i = 0

z = 1, z = - \frac{1}{2} + \frac{3}{2} i, z = - \frac{1}{2} - \frac{3}{2} i, z = \frac{1 + 3}{2} + i \frac{- 1 + 3}{2}, z = - 1 + i, z = \frac{1 - 3}{2} + i \frac{- 1 - 3}{2}

Solution steps

z^{6} - (3 + 2 i) z^{3} + 2 + 2 i = 0

Refine

(z^{6} - 3 z^{3} + 2) + i (2 - 2 z^{3}) = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

z^{6} - (3 + 2 i) z^{3} + 2 + 2 i = 0

Rewrite the equation with

u = z^{3}

and

u^{2} = z^{6}

u^{2} - (3 + 2 i) u + 2 + 2 i = 0

Solve

u^{2} - (3 + 2 i) u + 2 + 2 i = 0 : u = 1, u = 2 + 2 i

u = 1, u = 2 + 2 i

Substitute back

u = z^{3},

solve for

z

Solve

z^{3} = 1 : z = 1, z = - \frac{1}{2} + \frac{3}{2} i, z = - \frac{1}{2} - \frac{3}{2} i

Solve

z^{3} = 2 + 2 i : z = \frac{1 + 3}{2} + i \frac{- 1 + 3}{2}, z = - 1 + i, z = \frac{1 - 3}{2} + i \frac{- 1 - 3}{2}

The solutions are

z = 1, z = - \frac{1}{2} + \frac{3}{2} i, z = - \frac{1}{2} - \frac{3}{2} i, z = \frac{1 + 3}{2} + i \frac{- 1 + 3}{2}, z = - 1 + i, z = \frac{1 - 3}{2} + i \frac{- 1 - 3}{2}

16 x^{4} - x^{2} = 0

x^{3} - 3 x^{2} - 10 x + 24

\frac{Q ^{3}}{3} - 10 Q^{2} + 80 Q + 648 = 4 Q

x^{3} + x^{2} - 17 x + 15

x^{3} - x^{2} - 25 x + 25 = 0

What are the solutions to the equation z^6-(3+2i)z^3+2+2i=0 ?
The solutions to the equation z^6-(3+2i)z^3+2+2i=0 are
Find the zeros of z^6-(3+2i)z^3+2+2i=0
The zeros of z^6-(3+2i)z^3+2+2i=0 are