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

The answer to z^2+2iz^2+3iz=0 is z=0,z=-6/5-3/5 i

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z^2+2iz^2+3iz=0

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

z = 0, z = - \frac{6}{5} - \frac{3}{5} i

Solution steps

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

Substitute

z = x + y i

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

Expand

(x + y i)^{2} + 2 i (x + y i)^{2} + 3 i (x + y i) : (x^{2} - y^{2} - 4 x y - 3 y) + i (2 x^{2} - 2 y^{2} + 3 x + 2 x y)

(x^{2} - y^{2} - 4 x y - 3 y) + i (2 x^{2} - 2 y^{2} + 3 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 4 x y - 3 y) + i (2 x^{2} - 2 y^{2} + 3 x + 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} - 4 x y - 3 y = 0 2 x^{2} - 2 y^{2} + 3 x + 2 x y = 0]

[x^{2} - y^{2} - 4 x y - 3 y = 0 2 x^{2} - 2 y^{2} + 3 x + 2 x y = 0] : x = 0, x = - \frac{6}{5}, x = 0, x = - \frac{6}{5}, y = 0 y = - \frac{3}{5} y = 0 y = - \frac{3}{5}

Substitute back

z = x + y i

z = 0, z = - \frac{6}{5} - \frac{3}{5} i

(3 + 4 i) 2 - 2 (x - i y) = x + i y

(2 - 4 i) x - 4 y = 0

\frac{i}{z} = 1 + 1

\frac{1 - i}{1 + z \cdot i} = 2 - 3 i, z

z^{2} - (3 - 4 i) \cdot z + (2 - 8 i) = 0

What is the answer to z^2+2iz^2+3iz=0 ?
The answer to z^2+2iz^2+3iz=0 is z=0,z=-6/5-3/5 i