Inequality Proving: Complex Number

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We will discuss three methods of inequality proving in complex number.

 1. By Polar Form (Imp.)

 2. By Geometrical Method

 3. By Triangle's inequality
Ques: Show that \left| {\frac{z}{{|z|}} - 1} \right| \le |\arg z|
Sol: 1st method,

 Since, z = |z|(\cos \theta + i\sin \theta )

 or we can write, 

 \frac{z}{{|z|}} = (\cos \theta + i\sin \theta )

 or, \frac{z}{{|z|}} - 1 = (\cos \theta - 1) + i\sin \theta

Taking mod of both the sides,

 \left| {\frac{z}{{|z|}} - 1} \right| = \sqrt {{{(\cos \theta - 1)}^2} + {{\sin }^2}\theta }

or,           = \sqrt {2 - 2\cos \theta }


               = \sqrt {2(1 - \cos \theta )}

               = \sqrt {2.2{{\sin }^2}\frac{\theta }{2}}

or, \left| {\frac{z}{{|z|}} - 1} \right| = 2\left| {\sin \frac{\theta }{2}} \right|

 since, \left| {\sin x} \right| \le |x|

 therefore, \left| {\frac{z}{{|z|}} - 1} \right| \le 2.\frac{{\left| \theta \right|}}{2}

 or, \left| {\frac{z}{{|z|}} - 1} \right| \le |\arg z|   (hence proved)

 2nd Method,
Geometrical Method
{\frac{z}{{|z|}}} is a unit vector along z. We draw a unit circle centred at origin in above figure, so point P is (1,0).

 vector OA = \frac{z}{{|z|}}

 vector OA - vector OP = vector PA

 or, | vector OA - vector OP | = | vector PA |

 or, \left| {\frac{z}{{|z|}} - 1} \right| = |vecor PA|  \le arcPA

 since, arc = angle/ radius

 arcPA = |\theta |/1

 therefore, \left| {\frac{z}{{|z|}} - 1} \right| \le |\theta |

 or, \left| {\frac{z}{{|z|}} - 1} \right| \le |\arg z|

Ques: Show that |z - 1| \le ||z| - 1| + |z||\arg z|

Sol: |z - 1| = |z - |z| + |z| - 1|

 or, |z - 1| \le |z - |z|| + ||z| - 1| ( Triangle's inequality)

 Now, it is sufficient to prove, |z - |z|| \le |z||\arg z|

 or, \left| {\frac{z}{{|z|}} - 1} \right| \le |\arg z| (which has been proved in earlier question)
 2nd method,
Geometrical Method
OC = |z|
OP = 1
|PC|= ||z| - 1|

 In triangle APC,

 AP \le PC + AC

 |z - 1| \le ||z| - 1| + AC

 or, |z - 1| \le ||z| - 1| + {{\rm ar}\nolimits} cAC

 or, |z - 1| \le ||z| - 1| + |z||\arg z|  

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