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Let z1 and z2 be two complex numbers such that $${\overline z _1} = i{\overline z _2}$$ and $$\arg \left( {{{{z_1}} \over {{{\overline z }_2}}}} \right) = \pi $$. Then :

JEE · Math · previous-year question

  1. A.$$\arg {z_2} = {\pi \over 4}$$
  2. B.$$\arg {z_2} = - {{3\pi } \over 4}$$
  3. C.$$\arg {z_1} = {\pi \over 4}$$correct
  4. D.$$\arg {z_1} = - {{3\pi } \over 4}$$

Answer

C. $$\arg {z_1} = {\pi \over 4}$$

Explanation

$$\because$$ $${{{z_1}} \over {{z_2}}} = - i \Rightarrow {z_1} = - i{z_2}$$ $$ \Rightarrow \arg ({z_1}) = - {\pi \over 2} + \arg ({z_2})$$ ..... (i) Also $$\arg ({z_1}) - \arg ({\overline z _2}) = \pi $$ $$ \Rightarrow \arg ({z_1}) + \arg ({z_2}) = \pi $$ ..... (ii) From (i) and (ii), we get $$\arg ({z_1}) = {\pi \over 4}$$ and $$\arg ({z_2}) = {{3\pi } \over 4}$$

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