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Let n denote the number of solutions of the equation z2 + 3$$\overline z $$ = 0, where z is a complex number. Then the value of $$\sum\limits_{k = 0}^\infty {{1 \over {{n^k}}}} $$ is equal to :

JEE · Math · previous-year question

  1. A.1
  2. B.$${4 \over 3}$$correct
  3. C.$${3 \over 2}$$
  4. D.2

Answer

B. $${4 \over 3}$$

Explanation

z2 + 3$$\overline z $$ = 0 Put z = x + iy $$\Rightarrow$$ x2 $$-$$ y2 + 2ixy + 3(x $$-$$ iy) = 0 $$\Rightarrow$$ (x2 $$-$$ y2 + 3x) + i(2xy $$-$$ 3y) = 0 + i0 $$\therefore$$ x2 $$-$$ y2 + 3x = 0 ..... (1) 2xy $$-$$ 3y = 0 ..... (2) x = $${3 \over 2}$$, y = 0 Put x = $${3 \over 2}$$ in equation (1) $${9 \over 4} - {y^2} + {9 \over 2} = 0$$ $${y^2} = {{27} \over 4} \Rightarrow y = \pm {{3\sqrt 3 } \over 2}$$ $$\therefore$$ $$(x,y) = \left( {{3 \over 2},{{3\sqrt 3 } \over 2}} \right),\left( {{3 \over 2},{{ - 3\sqrt 3 } \over 2}} \right)$$ Put y = 0 $$\Rightarrow$$ x2 $$-$$ 0 + 3x = 0 x = 0, $$-$$3 $$\therefore$$ (x, y) = (0, 0), ($$-$$3, 0) $$\therefore$$ No of solutions = n = 4 $$\sum\limits_{K = 0}^\infty {\left( {{1 \over {{n^k}}}} \right)} = \sum\limits_{K = 0}^\infty {\left( {{1 \over {4{n^k}}}} \right)} $$ $$ = {1 \over 1} + {1 \over 4} + {1 \over {16}} + {1 \over {64}} + ......$$ $$ = {1 \over {1 - {1 \over 4}}} = {4 \over 3}$$

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