If |z^2-1|=|z|^2+1, then z lies on

Question

If $|z^2-1|=|z|^2+1$, then $z$ lies on

Accepted Answer

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$

Solutions · 1