(z+3i+2)(2-i)=z

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Solution for (z+3i+2)(2-i)=z equation: 


Simplifying
(z + 3i + 2)(2 + -1i) = z

Reorder the terms:
(2 + 3i + z)(2 + -1i) = z

Multiply (2 + 3i + z) * (2 + -1i)
(2(2 + -1i) + 3i * (2 + -1i) + z(2 + -1i)) = z
((2 * 2 + -1i * 2) + 3i * (2 + -1i) + z(2 + -1i)) = z
((4 + -2i) + 3i * (2 + -1i) + z(2 + -1i)) = z
(4 + -2i + (2 * 3i + -1i * 3i) + z(2 + -1i)) = z
(4 + -2i + (6i + -3i2) + z(2 + -1i)) = z
(4 + -2i + 6i + -3i2 + (2 * z + -1i * z)) = z

Reorder the terms:
(4 + -2i + 6i + -3i2 + (-1iz + 2z)) = z
(4 + -2i + 6i + -3i2 + (-1iz + 2z)) = z

Reorder the terms:
(4 + -2i + 6i + -1iz + -3i2 + 2z) = z

Combine like terms: -2i + 6i = 4i
(4 + 4i + -1iz + -3i2 + 2z) = z

Solving
4 + 4i + -1iz + -3i2 + 2z = z

Solving for variable 'i'.

Combine like terms: 2z + -1z = 1z
4 + 4i + -1iz + -3i2 + 1z = z + -1z

Combine like terms: z + -1z = 0
4 + 4i + -1iz + -3i2 + 1z = 0

The solution to this equation could not be determined.

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