(z^2-2iz+3)(2iz+1)=0

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Solution for (z^2-2iz+3)(2iz+1)=0 equation:


Simplifying
(z2 + -2iz + 3)(2iz + 1) = 0

Reorder the terms:
(3 + -2iz + z2)(2iz + 1) = 0

Reorder the terms:
(3 + -2iz + z2)(1 + 2iz) = 0

Multiply (3 + -2iz + z2) * (1 + 2iz)
(3(1 + 2iz) + -2iz * (1 + 2iz) + z2(1 + 2iz)) = 0
((1 * 3 + 2iz * 3) + -2iz * (1 + 2iz) + z2(1 + 2iz)) = 0
((3 + 6iz) + -2iz * (1 + 2iz) + z2(1 + 2iz)) = 0
(3 + 6iz + (1 * -2iz + 2iz * -2iz) + z2(1 + 2iz)) = 0
(3 + 6iz + (-2iz + -4i2z2) + z2(1 + 2iz)) = 0
(3 + 6iz + -2iz + -4i2z2 + (1 * z2 + 2iz * z2)) = 0

Reorder the terms:
(3 + 6iz + -2iz + -4i2z2 + (2iz3 + 1z2)) = 0
(3 + 6iz + -2iz + -4i2z2 + (2iz3 + 1z2)) = 0

Reorder the terms:
(3 + 6iz + -2iz + 2iz3 + -4i2z2 + 1z2) = 0

Combine like terms: 6iz + -2iz = 4iz
(3 + 4iz + 2iz3 + -4i2z2 + 1z2) = 0

Solving
3 + 4iz + 2iz3 + -4i2z2 + 1z2 = 0

Solving for variable 'i'.

The solution to this equation could not be determined.

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