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

The answer to x^2+2x+1-i=0 is x=-(-sqrt(2)+2)/2+1/(sqrt(2))i,x=-(sqrt(2)+2)/2-1/(sqrt(2))i

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

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

x = - \frac{- 2 + 2}{2} + \frac{1}{2} i, x = - \frac{2 + 2}{2} - \frac{1}{2} i

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

x = a + bi

x = - \frac{- 2 + 2}{2} + \frac{1}{2} i, x = - \frac{2 + 2}{2} - \frac{1}{2} i

y^{2} = 1 - i

∣ z - (1 - i) ∣ = 6

m = p [(1 + i) t - 1] i

a - bi = - b + ai

(3 - 4 i) (a + bi) = 25

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