What is the answer to iz^2+2z+8-i=0 ?

The answer to iz^2+2z+8-i=0 is z=2+3i,z=-2-i

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iz^2+2z+8-i=0

i z^{2} + 2 z + 8 - i = 0

z = 2 + 3 i, z = - 2 - i

Solution steps

i z^{2} + 2 z + 8 - i = 0

Substitute

z = x + y i

i (x + y i)^{2} + 2 (x + y i) + 8 - i = 0

Expand

i (x + y i)^{2} + 2 (x + y i) + 8 - i : (- 2 x y + 2 x + 8) + i (- 1 + x^{2} - y^{2} + 2 y)

(- 2 x y + 2 x + 8) + i (- 1 + x^{2} - y^{2} + 2 y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(- 2 x y + 2 x + 8) + i (- 1 + x^{2} - y^{2} + 2 y) = 0 + 0 i

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

[- 2 x y + 2 x + 8 = 0 - 1 + x^{2} - y^{2} + 2 y = 0]

[- 2 x y + 2 x + 8 = 0 - 1 + x^{2} - y^{2} + 2 y = 0] : (x = 2, x = - 2, y = 3 y = - 1)

Substitute back

z = x + y i

z = 2 + 3 i, z = - 2 - i

i = \frac{x}{a x + b}, x

(1 + i) (a + bi) + 5 = 2 \cdot 3 i

r = 2 t i + (t^{2} - 1) j

\frac{z - 1}{z} = 5 + 2 i

- 6 i = \frac{- i}{120 π \cdot C}

What is the answer to iz^2+2z+8-i=0 ?
The answer to iz^2+2z+8-i=0 is z=2+3i,z=-2-i