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If $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$, then z lies on :

JEE · Math · previous-year question

  1. A.an ellipse
  2. B.the imaginary axiscorrect
  3. C.a circle
  4. D.the real axis

Answer

B. the imaginary axis

Explanation

Given $$\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1$$, By squaring both sides we get, $${\left| {{z^2} - 1} \right|^2}$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$ $$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\overline {\left( {{z^2} - 1} \right)} $$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$ [ as $${{{\left| z \right|}^2}}$$ = $$z\overline z $$ ] $$ \Rightarrow $$ $$\left( {{z^2} - 1} \right)$$$$\left( {{{\left( {\overline z } \right)}^2} - 1} \right)$$ = $${\left( {{{\left| z \right|}^2} + 1} \right)^2}$$ $$ \Rightarrow $$ $${\left( {z\overline z } \right)^2}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1 $$ \Rightarrow $$ $${\left| z \right|^4}$$ $$-$$ $${{z^2}}$$ $$-$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 1 = $${\left| z \right|^4}$$ $$+$$ 2$${{{\left| z \right|}^2}}$$ $$+$$ 1 $$ \Rightarrow $$ $${{z^2}}$$ $$+$$$${{{\left( {\overline z } \right)}^2}}$$ $$+$$ 2$${z\overline z }$$ = 0 $$ \Rightarrow $$ $${\left( {z + \overline z } \right)^2}$$ = 0 $$ \Rightarrow $$ $${z + \overline z }$$ = 0 $$ \Rightarrow $$ $$z$$ = $$-$$ $${\overline z }$$ If $$z$$ = x + iy then $${\overline z }$$ = x - iy $$\therefore$$ x + iy = - (x - iy) $$ \Rightarrow $$ x + iy = - x + iy $$ \Rightarrow $$ x = 0 $$\therefore$$ z is purely imaginary. So, it is lie on the imaginary axis.

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