What is the answer to solvefor z,z^2-(1-4i)z-4i=0 ?

The answer to solvefor z,z^2-(1-4i)z-4i=0 is z=-4i,z=1

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solvefor z,z^2-(1-4i)z-4i=0

solve for

z, z^{2} - (1 - 4 i) z - 4 i = 0

z = - 4 i, z = 1

Solution steps

z^{2} - (1 - 4 i) z - 4 i = 0

Substitute

z = x + y i

(x + y i)^{2} - (1 - 4 i) (x + y i) - 4 i = 0

Expand

(x + y i)^{2} - (1 - 4 i) (x + y i) - 4 i : (x^{2} - y^{2} - x - 4 y) + i (- 4 - y + 4 x + 2 x y)

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[x^{2} - y^{2} - x - 4 y = 0 - 4 - y + 4 x + 2 x y = 0] : (x = 0, x = 1, y = - 4 y = 0)

Substitute back

z = x + y i

z = - 4 i, z = 1

x^{2} + 2 i x - 1 = 0

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

\frac{x + i y}{1 + x + i y} = 3 + 4 i

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

3 x - 2 y i = 6 + 4 i

What is the answer to solvefor z,z^2-(1-4i)z-4i=0 ?
The answer to solvefor z,z^2-(1-4i)z-4i=0 is z=-4i,z=1