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Question

If $$\left| z \right| \le 1,\ \left| w \right| \le 1$$ show that
$${ \left| z-w \right|  }^{ 2 }\le { \left( \left| z \right| -\left| w \right|  \right)  }^{ 2 }={ \left( Arg\ z-Arg\ w \right)  }^{ 2 }$$


Solution

Let $$z= re^{ i\theta  }$$, $$w=Re^{ i\phi  }\quad \therefore\ r\le 1$$, $$R\le 1$$,
$$Arg\ z=\theta,\ Arg\ w=\phi$$
$$\left| z-w \right| ^{ 2 }={ r }^{ 2 }+{ R }^{ 2 }-2rR\cos { \left( \theta -\phi  \right)  }$$
$$r^{ 2 }+R^{ 2 }-2rR+2rR[1-\cos(\theta-\phi)]$$
$$=\left( r-R \right) ^{ 2 }+2rR.2\sin { ^{ 2 } } \left\{ \dfrac { \theta -\phi  }{ 2 }  \right\}$$
$$\le \left( r-R \right) ^{ 2 }+4.1.1\left( \dfrac { \theta -\phi  }{ 2 }  \right) ^{ 2 }$$
$$=\left( \left| z \right| -\left| w \right|  \right) ^{ 2 }+\left( \ Arg\ z-\ Arg\ w \right) ^{ 2 }$$
$$R<1$$, $$\sin { \psi  } <\psi$$ where $$\psi$$ is a $$+$$ive angle. Also $$\sin ^{ 2 } \psi <\psi ^{ 2 }$$.

Mathematics

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