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

The answer to (x+1)(x+i)=0 is x=-i,x=-1

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

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

x = - i, x = - 1

Solution steps

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

Substitute

x = a + bi

(a + bi + 1) (a + bi + i) = 0

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

x = a + bi

x = - i, x = - 1

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

i (x) = - 0.01 x^{2} + 10 x + 300

z, z^{2} = 5 - 12 i

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

- 3 x - 7 y i = 9 - 21 i