InterviewSolution
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InterviewSolution
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Solve the equation `|z|=z+1+2idot` |
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Answer» The given equation is `|z| = z + 1 + 2i " " ...(i) ` Let `z = x + iy` From Eq. (i). |x + iy| = x + iy + 1 + 2i` `rArr sqrt(x^(2)+Y^(2) = x + iy + 1 + 2i)" " [:. |z| + sqrt (x^(2)+Y^(2))]` `rArr sqrt(x^(2)+Y^(2)) = (x + 1) + i (y + 2)` On squaring both sides, we get `x^(2) + y^(2) = (x+ 1)^(2) + i^(2) (y +2)^(2) + 2i(x + 1)(y +2)` `rArr x^(2) + y^(2) = x^(2) +2x + 1- y ^(2) - 4y -4 + 2i(x + 1)(y +2)` On comparing real and imaginary parts , `x^(2) + y^(2) = x^(2) +2x + 1- y ^(2) - 4y -4` i. e., `2y^(2) = 2x - 4y - 3` ....(ii) and `2(x + 1)(y + 2) 0` `(x + 1) = 0 or (y + 2) = 0` `rArr x = - 1 or y = - 2` For ` x = - 1` we get, `2y^(2) = -2- 4 y 3` `2y^(2) = + 4 + 5 = 0 [using Eq. (ii)]` `rArr y = (- 4pmsqrt(16 - 2 xx 4 xx 5))/(4)` `rArr y = (- 4pm(sqrt(-24)))/(4)notinR` Now, for y = - 2, Then, `2(-2)^(2) = 2x - 4 (-2) -3 [using Eq. (ii)]` `8 = 2x + 8 - 3` `2x = 3 rArr x = 3//2` `z = x + iy = 3/2 - 2i` |
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