(1+3i)*z+(2-i)=0

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Solution for (1+3i)*z+(2-i)=0 equation:


Simplifying
(1 + 3i) * z + (2 + -1i) = 0

Reorder the terms for easier multiplication:
z(1 + 3i) + (2 + -1i) = 0
(1 * z + 3i * z) + (2 + -1i) = 0

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

Remove parenthesis around (2 + -1i)
3iz + 1z + 2 + -1i = 0

Reorder the terms:
2 + -1i + 3iz + 1z = 0

Solving
2 + -1i + 3iz + 1z = 0

Solving for variable 'i'.

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

Add '-2' to each side of the equation.
2 + -1i + 3iz + -2 + 1z = 0 + -2

Reorder the terms:
2 + -2 + -1i + 3iz + 1z = 0 + -2

Combine like terms: 2 + -2 = 0
0 + -1i + 3iz + 1z = 0 + -2
-1i + 3iz + 1z = 0 + -2

Combine like terms: 0 + -2 = -2
-1i + 3iz + 1z = -2

Add '-1z' to each side of the equation.
-1i + 3iz + 1z + -1z = -2 + -1z

Combine like terms: 1z + -1z = 0
-1i + 3iz + 0 = -2 + -1z
-1i + 3iz = -2 + -1z

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

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

Combine like terms: -2 + 2 = 0
2 + -1i + 3iz + z = 0 + -1z + z
2 + -1i + 3iz + z = -1z + z

Combine like terms: -1z + z = 0
2 + -1i + 3iz + z = 0

The solution to this equation could not be determined.

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