Geometrical Meaning of Argument and Modulus - II: Complex Number

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 13. {{\rm Re}\nolimits} (z) \ge \frac{1}{2} 

z = x + iy

then, x \ge \frac{1}{2}

14. {{\rm Re}\nolimits} \left( {\frac{1}{z}} \right) \le \frac{1}{2}

if z = x + iy

1/z = 1/ (x + iy)

      = \frac{{x - iy}}{{(x + iy)(x - iy)}}

      = \frac{{(x - iy)}}{{{x^2} + {y^2}}}

        = \frac{x}{{{x^2} + {y^2}}} - i\frac{y}{{{x^2} + {y^2}}}

since, {{\rm Re}\nolimits} \left( {\frac{1}{z}} \right) \le \frac{1}{2}

therefore, \frac{x}{{{x^2} + {y^2}}} \le \frac{1}{2}

or, {{x^2} + {y^2} - 2x \ge 0}

 is the equation of circle having centre (0,1) and radius '1'. The region represented of {{x^2} + {y^2} - 2x \ge 0} is outside of the circle.

centre \equiv (1,0), radius = 1



NOTE:
|{z_1} - {z_2}| = distance between {z_1} & {z_2} .

 1. | z + i | = | z - i |

 i.e. | z + i | = | z - (-i ) | distance of z from '-i' 
and | z - i | = distance of z from 'i' 

 Therefore, locus of 'z' is the perpendicular bisector of the line joining 'i' and '-i'



2. | z + 2 - 3i | = | z - 3 + i |

 | z + 2 - 3i | = distance of z from -2 + 3i 
| z - 3 + i | = distance of z from 3 - i

 locus of z is the perpendicular bisector of line joining (-2,3) and (3,-1)


3. | z - i | = 1

 distance of z from i = 1 ( constant)

 therefore the equation represents circle 

 centre \equiv (0,1), Radius = 1
Equation in coordinate form:
{x^2} + {(y - 1)^2} = 1
4. 1 < | z - 2 + 3i | < 2

 or, 1 < | z - ( 2 - 3i ) | < 2

 centre \equiv (2, - 3)



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