What is the answer to z^2+(1-2i)z+1+5i=0 ?

The answer to z^2+(1-2i)z+1+5i=0 is z=1-i,z=-2+3i

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z^2+(1-2i)z+1+5i=0

z^{2} + (1 - 2 i) z + 1 + 5 i = 0

z = 1 - i, z = - 2 + 3 i

Solution steps

z^{2} + (1 - 2 i) z + 1 + 5 i = 0

Substitute

z = x + y i

(x + y i)^{2} + (1 - 2 i) (x + y i) + 1 + 5 i = 0

Expand

(x + y i)^{2} + (1 - 2 i) (x + y i) + 1 + 5 i : (x^{2} - y^{2} + x + 2 y + 1) + i (5 + y - 2 x + 2 x y)

(x^{2} - y^{2} + x + 2 y + 1) + i (5 + y - 2 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} + x + 2 y + 1) + i (5 + y - 2 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 + 2 y + 1 = 0 5 + y - 2 x + 2 x y = 0]

[x^{2} - y^{2} + x + 2 y + 1 = 0 5 + y - 2 x + 2 x y = 0] : (x = 1, x = - 2, y = - 1 y = 3)

Substitute back

z = x + y i

z = 1 - i, z = - 2 + 3 i

a^{2} + 2 abi - b^{2} = 2 i

a + (b + 1) \cdot i = 4

0.52 + 0.32 bi = (a + b) - 3 i

A i = A - P

x, x^{2} = 3 + i

What is the answer to z^2+(1-2i)z+1+5i=0 ?
The answer to z^2+(1-2i)z+1+5i=0 is z=1-i,z=-2+3i