InterviewSolution
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InterviewSolution
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If `|z+1|=z+2(1+i),`find `zdot` |
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Answer» Given that, `|z +1| = z + 2 ( + i) ... (i)` `z = x + iy` then, `|x + iy + 1| = x + iy + 2 (1 + i)` `rArr |x + 1 + iy| = (x + 2) + i (y + 2)` `rArr" sqrt((x + 1)^(2) + y^(2) = (x + 2) + i (y + 2))` On squaring both sides, we get `(x + 1)^(2) + y^(2) = (x + 2)^(2) +i ^(2)(y + 2)^(2) +2i (x + 2)(y +2)` `rArr x^(2) + 2x + 1 + y^(2) = x^(2) + 4x + 4 - y^(2) - 4y - 4 +2i(x + 2)(y + 2)` `rArr x^(2) + y^(2)+ 2x + 1 = x^(2) - y^(2) + 4x - 4y + 2i(x + 2)(y + 2)` On comparing real and imaginary parts, we get ` x^(2) + y^(2) + 2x + 1 = x^(2) - y^(2) + 4x -4y` `rArr 2y^(2) - 2x + 4y + 1 = 0 ...(ii)` and `2(x + 2) (y + 2) =0` `rArr x + 2 = 0 or y + 2 = 0 ` `x = - 2 or y = - 2 ...(iii)` For `x = - 2, 2y^(2) + 4 + 4y + 1 = 0 [ using Eq. (ii)]` `rArr 2y^(2) + 4y + 5 =` `rArr 16 - 4 xx 2 xx 5 lt0 ` `:. "Discriminant" , D = b^(2) - 4ac lt 0 ` `rArr 2y^(2) + 4y + 5 "has no real roots"`. For y = -2, `2(-)6(2) - 2x + 4 (-2) + 1 = 0 [ using Eq. (ii)]` `rArr 8 - 2x - 8 + 1 = 0` `rArr x = 1//2` `:. " " z = x + iy = (1)/(2) - 2i` |
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