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

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

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

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

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

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

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

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

i (x) = - 100 x^{2} + 1000 x + 7500

- w^{2} + 9 i w + 20 = 0

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

3 x + 21 i = 6 + 7

\frac{1 + ai}{6 + bi} = 3 + 7 i

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