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

The answer to x^2+(1-i)x+i=0 is x=0.36602…-0.36602…i,x=-1.36602…+1.36602…i

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

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

x = 0.36602 \dots - 0.36602 \dots i, x = - 1.36602 \dots + 1.36602 \dots i

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[a^{2} - b^{2} + a + b = 0 1 - a + b + 2 ab = 0] : (a = 0.36602 \dots, a = - 1.36602 \dots, b = - 0.36602 \dots b = 1.36602 \dots)

Substitute back

x = a + bi

x = 0.36602 \dots - 0.36602 \dots i, x = - 1.36602 \dots + 1.36602 \dots i

∣ z - i ∣ = 3

∣ z - i ∣ = 4

(1 + i) x - (1 - 2 i) y = 1 + 4 i

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

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

What is the answer to x^2+(1-i)x+i=0 ?
The answer to x^2+(1-i)x+i=0 is x=0.36602…-0.36602…i,x=-1.36602…+1.36602…i