(2z+1-i)(iz+1)=0

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


Simplifying
(2z + 1 + -1i)(iz + 1) = 0

Reorder the terms:
(1 + -1i + 2z)(iz + 1) = 0

Reorder the terms:
(1 + -1i + 2z)(1 + iz) = 0

Multiply (1 + -1i + 2z) * (1 + iz)
(1(1 + iz) + -1i * (1 + iz) + 2z * (1 + iz)) = 0
((1 * 1 + iz * 1) + -1i * (1 + iz) + 2z * (1 + iz)) = 0
((1 + 1iz) + -1i * (1 + iz) + 2z * (1 + iz)) = 0
(1 + 1iz + (1 * -1i + iz * -1i) + 2z * (1 + iz)) = 0
(1 + 1iz + (-1i + -1i2z) + 2z * (1 + iz)) = 0
(1 + 1iz + -1i + -1i2z + (1 * 2z + iz * 2z)) = 0

Reorder the terms:
(1 + 1iz + -1i + -1i2z + (2iz2 + 2z)) = 0
(1 + 1iz + -1i + -1i2z + (2iz2 + 2z)) = 0

Reorder the terms:
(1 + -1i + 1iz + 2iz2 + -1i2z + 2z) = 0
(1 + -1i + 1iz + 2iz2 + -1i2z + 2z) = 0

Solving
1 + -1i + 1iz + 2iz2 + -1i2z + 2z = 0

Solving for variable 'i'.

The solution to this equation could not be determined.

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