z^2+(1+2i)z+1=0

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


Simplifying
z2 + (1 + 2i) * z + 1 = 0

Reorder the terms for easier multiplication:
z2 + z(1 + 2i) + 1 = 0
z2 + (1 * z + 2i * z) + 1 = 0

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

Reorder the terms:
1 + 2iz + 1z + z2 = 0

Solving
1 + 2iz + 1z + z2 = 0

Solving for variable 'i'.

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

Add '-1' to each side of the equation.
1 + 2iz + 1z + -1 + z2 = 0 + -1

Reorder the terms:
1 + -1 + 2iz + 1z + z2 = 0 + -1

Combine like terms: 1 + -1 = 0
0 + 2iz + 1z + z2 = 0 + -1
2iz + 1z + z2 = 0 + -1

Combine like terms: 0 + -1 = -1
2iz + 1z + z2 = -1

Add '-1z' to each side of the equation.
2iz + 1z + -1z + z2 = -1 + -1z

Combine like terms: 1z + -1z = 0
2iz + 0 + z2 = -1 + -1z
2iz + z2 = -1 + -1z

Add '-1z2' to each side of the equation.
2iz + z2 + -1z2 = -1 + -1z + -1z2

Combine like terms: z2 + -1z2 = 0
2iz + 0 = -1 + -1z + -1z2
2iz = -1 + -1z + -1z2

Divide each side by '2z'.
i = -0.5z-1 + -0.5 + -0.5z

Simplifying
i = -0.5z-1 + -0.5 + -0.5z

Reorder the terms:
i = -0.5 + -0.5z-1 + -0.5z

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