What is the answer to z^2+6iz-1=0 ?

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

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z^2+6iz-1=0

z^{2} + 6 i z - 1 = 0

z = (- 3 - 22) i, z = (- 3 + 22) i

Solution steps

z^{2} + 6 i z - 1 = 0

Substitute

z = x + y i

(x + y i)^{2} + 6 i (x + y i) - 1 = 0

Expand

(x + y i)^{2} + 6 i (x + y i) - 1 : (x^{2} - y^{2} - 6 y - 1) + i (6 x + 2 x y)

(x^{2} - y^{2} - 6 y - 1) + i (6 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 6 y - 1) + i (6 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} - 6 y - 1 = 0 6 x + 2 x y = 0]

[x^{2} - y^{2} - 6 y - 1 = 0 6 x + 2 x y = 0] : (x = 0, x = 0, y = - 3 - 22 y = - 3 + 22)

Substitute back

z = x + y i

z = (- 3 - 22) i, z = (- 3 + 22) i

6 + \frac{1}{x} = 6, x^{2} + x^{- 2}

i (x) = 5 x - x^{2}

z^{2} - (5 + i) z + (8 + i) = 0

(2 x + 1) + (5 - y) i = - 3 + 2 i

x^{2} + 6 (3 - i) - 26 = 0

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