What is the answer to z^2+8-i6=0 ?

The answer to z^2+8-i6=0 is z=1+3i,z=-1-3i

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z^2+8-i6=0

z^{2} + 8 - i 6 = 0

z = 1 + 3 i, z = - 1 - 3 i

Solution steps

z^{2} + 8 - i 6 = 0

Substitute

z = x + y i

(x + y i)^{2} + 8 - i 6 = 0

Expand

(x + y i)^{2} + 8 - i 6 : (x^{2} - y^{2} + 8) + i (- 6 + 2 x y)

(x^{2} - y^{2} + 8) + i (- 6 + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} + 8) + i (- 6 + 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} + 8 = 0 - 6 + 2 x y = 0]

[x^{2} - y^{2} + 8 = 0 - 6 + 2 x y = 0] : (x = 1, x = - 1, y = 3 y = - 3)

Substitute back

z = x + y i

z = 1 + 3 i, z = - 1 - 3 i

\frac{5 + i 2}{( - 1 + i 6 ) x} = - 1

5 i (x + i y) + 3 (x - i y) + i y = - 4 - 5 i

f, i f = M

- \frac{1}{z - 3 i} + 1 - z = 0

x, z = \frac{1 - x i}{1 + x i}