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

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


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

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

Reorder the terms:
-6i + 2iz + 2z = z(1 + i)
-6i + 2iz + 2z = (1 * z + i * z)

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

Solving
-6i + 2iz + 2z = iz + 1z

Solving for variable 'i'.

Move all terms containing i to the left, all other terms to the right.

Add '-1iz' to each side of the equation.
-6i + 2iz + -1iz + 2z = iz + -1iz + 1z

Combine like terms: 2iz + -1iz = 1iz
-6i + 1iz + 2z = iz + -1iz + 1z

Combine like terms: iz + -1iz = 0
-6i + 1iz + 2z = 0 + 1z
-6i + 1iz + 2z = 1z

Add '-2z' to each side of the equation.
-6i + 1iz + 2z + -2z = 1z + -2z

Combine like terms: 2z + -2z = 0
-6i + 1iz + 0 = 1z + -2z
-6i + 1iz = 1z + -2z

Combine like terms: 1z + -2z = -1z
-6i + 1iz = -1z

Combine like terms: -1z + z = 0
-6i + 1iz + z = 0

The solution to this equation could not be determined.

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