What is the answer to x^2+x(1+i)+1=0 ?

The answer to x^2+x(1+i)+1=0 is x=-0.25706…+0.52908…i,x=-0.74293…-1.52908…i

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x^2+x(1+i)+1=0

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

x = - 0.25706 \dots + 0.52908 \dots i, x = - 0.74293 \dots - 1.52908 \dots i

Solution steps

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

Substitute

x = a + bi

(a + bi)^{2} + (a + bi) (1 + i) + 1 = 0

Expand

(a + bi)^{2} + (a + bi) (1 + i) + 1 : (a^{2} - b^{2} + a - b + 1) + i (a + b + 2 ab)

(a^{2} - b^{2} + a - b + 1) + i (a + b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + a - b + 1) + i (a + b + 2 ab) = 0 + 0 i

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

[a^{2} - b^{2} + a - b + 1 = 0 a + b + 2 ab = 0]

[a^{2} - b^{2} + a - b + 1 = 0 a + b + 2 ab = 0] : (a = - 0.25706 \dots, a = - 0.74293 \dots, b = 0.52908 \dots b = - 1.52908 \dots)

Substitute back

x = a + bi

x = - 0.25706 \dots + 0.52908 \dots i, x = - 0.74293 \dots - 1.52908 \dots i

(3 a + 2 i) - (5 - 2 bi) = 2 - 6 i

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

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

\frac{( x + i y )}{( x - i y )} = - i

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

What is the answer to x^2+x(1+i)+1=0 ?
The answer to x^2+x(1+i)+1=0 is x=-0.25706…+0.52908…i,x=-0.74293…-1.52908…i