(1+i)^2+2+2i/2+i

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Solution for (1+i)^2+2+2i/2+i equation: 


i in (-oo:+oo)

(i+1)^2+(2*i)/2+i+2 = 0

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

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

i^2+3*i+i+1+2 = 0

i^2+4*i+1+2 = 0

i^2+4*i+3 = 0

i^2+4*i+3 = 0

i^2+4*i+3 = 0

DELTA = 4^2-(1*3*4)

DELTA = 4

DELTA > 0

i = (4^(1/2)-4)/(1*2) or i = (-4^(1/2)-4)/(1*2)

i = -1 or i = -3

(i+3)*(i+1) = 0

(i+3)*(i+1) = 0

( i+1 )

i+1 = 0 // - 1

i = -1

( i+3 )

i+3 = 0 // - 3

i = -3

i in { -1, -3 }

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