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

The answer to 3z^2-2+z-3iz^2+2i+zi=0 is z=(sqrt(23))/6-1/6 i,z=-(sqrt(23))/6-1/6 i

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3z^2-2+z-3iz^2+2i+zi=0

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

z = \frac{23}{6} - \frac{1}{6} i, z = - \frac{23}{6} - \frac{1}{6} i

Solution steps

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

Substitute

z = x + y i

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

Expand

3 (x + y i)^{2} - 2 + x + y i - 3 i (x + y i)^{2} + 2 i + (x + y i) i : (3 x^{2} - 3 y^{2} - 2 + x + 6 x y - y) + i (2 + x + y - 3 x^{2} + 3 y^{2} + 6 x y)

(3 x^{2} - 3 y^{2} - 2 + x + 6 x y - y) + i (2 + x + y - 3 x^{2} + 3 y^{2} + 6 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(3 x^{2} - 3 y^{2} - 2 + x + 6 x y - y) + i (2 + x + y - 3 x^{2} + 3 y^{2} + 6 x y) = 0 + 0 i

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

[3 x^{2} - 3 y^{2} - 2 + x + 6 x y - y = 0 2 + x + y - 3 x^{2} + 3 y^{2} + 6 x y = 0]

[3 x^{2} - 3 y^{2} - 2 + x + 6 x y - y = 0 2 + x + y - 3 x^{2} + 3 y^{2} + 6 x y = 0] : (x = \frac{23}{6}, x = - \frac{23}{6}, y = - \frac{1}{6} y = - \frac{1}{6})

Substitute back

z = x + y i

z = \frac{23}{6} - \frac{1}{6} i, z = - \frac{23}{6} - \frac{1}{6} i

∣ z + 3 ∣ = 1 - i z

2 x + 8 i = 6 - (2 y) i

9 x + (y) i - 5 = - 12 i + 4

i = (a + bi)^{2}

\frac{1 + ai}{5 + bi} = 2 + 7 i

What is the answer to 3z^2-2+z-3iz^2+2i+zi=0 ?
The answer to 3z^2-2+z-3iz^2+2i+zi=0 is z=(sqrt(23))/6-1/6 i,z=-(sqrt(23))/6-1/6 i