What is the answer to z^2+3z+2i=0 ?

The answer to z^2+3z+2i=0 is z=0.12178…-0.61660…i,z=-3.12178…+0.61660…i

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z^2+3z+2i=0

z^{2} + 3 z + 2 i = 0

z = 0.12178 \dots - 0.61660 \dots i, z = - 3.12178 \dots + 0.61660 \dots i

Solution steps

z^{2} + 3 z + 2 i = 0

Substitute

z = x + y i

(x + y i)^{2} + 3 (x + y i) + 2 i = 0

Expand

(x + y i)^{2} + 3 (x + y i) + 2 i : (x^{2} - y^{2} + 3 x) + i (2 + 3 y + 2 x y)

(x^{2} - y^{2} + 3 x) + i (2 + 3 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} + 3 x) + i (2 + 3 y + 2 x y) = 0 + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[x^{2} - y^{2} + 3 x = 0 2 + 3 y + 2 x y = 0]

[x^{2} - y^{2} + 3 x = 0 2 + 3 y + 2 x y = 0] : (x = 0.12178 \dots, x = - 3.12178 \dots, y = - 0.61660 \dots y = 0.61660 \dots)

Substitute back

z = x + y i

z = 0.12178 \dots - 0.61660 \dots i, z = - 3.12178 \dots + 0.61660 \dots i

(2 i + 1)^{2} = (m + ni) (2 - i)

2 (x - y i) + 5 (x + y i) + (x + y i)^{2} = 0

- 0.008 i = \frac{1}{x \cdot i}

M = C (1 + i)

5 (a + bi) = (a - bi) (- 3 + 4 i)

What is the answer to z^2+3z+2i=0 ?
The answer to z^2+3z+2i=0 is z=0.12178…-0.61660…i,z=-3.12178…+0.61660…i