Some Solved questions on Complex Number

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1. If  |z| = min { |z - 1| , |z + 1| } then

(a) |z + \overline z | = 1/2
(b) z + \overline z = 1
(c) |z + \overline {z|} = 1
(d) None of the above

Sol: |z - 1| = distance of z from '1'

|z + 1| = distance of z from '-1'

and |z - 0| = |z| = distance of z from '0'

The above equation  |z| = min { |z - 1| , |z + 1| } represents,

either |z| = |z - 1| , when |z - 1| < |z + 1|

or |z| = |z + 1|, when |z + 1| < |z - 1|

i.e. whichever is minimum in |z - 1| and |z + 1|, |z| is equal to that.

When |z| = |z - 1|

i.e. distance of 'z' from '0' = distance of 'z' from 1

implies 'z' has to be on the perpendicular bisector of the line joining the point 0 and 1.

the equation of the perpendicular bisector, x = 1/2

similarly, when |z| = |z + 1|

implies 'z' has to be on the perpendicular bisector of the line joining the point 0 and 1.

the equation of the perpendicular bisector, x = -1/2

   It is clear from the above figure that when z is on x = 1/2 i.e.                  |z| = |z - 1| then z + \overline z = 1

 and when z is on x = -1/2 i.e. |z| = |z + 1| then z + \overline z = - 1

 in both the condition, |z + \overline {z|} = 1 

 Therefore option (c) is correct 

 Also remember the relation shown below in z , -z and \overline z  , 
2. If |{z_1} + {z_2}| = |{z_1} - {z_2}| then ,

 a) |\arg {z_1} - \arg {z_2}| = \frac{\pi }{2}
b) |\arg {z_1} - \arg {z_2}| = \pi
c) \frac{{{z_1}}}{{{z_2}}} \Rightarrow  Purely Real
d) \frac{{{z_1}}}{{{z_2}}} \Rightarrow  Purely Imaginary

 Sol: Remember the vector representation of complex number, therefore addition of complex number is represented by one diagonal of Parallelogram and subtraction is represented by other diagonal.

 but if |{z_1} + {z_2}| = |{z_1} - {z_2}| i.e. both the diagonals are equal than the parallelogram must be a rectangle or square.


therefore, angle between {z_1} and {z_2} is pi/2

 => |\arg {z_1} - \arg {z_2}| = \frac{\pi }{2}

 therefore, option (a) is correct.

 Further, 
since, |{z_1} + {z_2}| = |{z_1} - {z_2}|

 \Rightarrow |{z_2}|\left| {\frac{{{z_1}}}{{{z_2}}} + 1} \right| = |{z_2}|\left| {\frac{{{z_1}}}{{{z_2}}} - 1} \right|

 or, \left| {\frac{{{z_1}}}{{{z_2}}} + 1} \right| = \left| {\frac{{{z_1}}}{{{z_2}}} - 1} \right|

 let, {\frac{{{z_1}}}{{{z_2}}} = w}

 therefore, \left| {w + 1} \right| = \left| {w - 1} \right| 
(represents the equation of perpendicular bisector of line joining the point 1 and -1.)

 since w is at equal distance from 1 and -1 

 therefore w must be on perpendicular bisector of line joining 1 and -1 i.e. w lies on imaginary or y-axis.

 i.e. |arg.w| = pi/2

 therefore, w must be Purely imaginary
or, \frac{{{z_1}}}{{{z_2}}} \Rightarrow  Purely Imaginary

 OR, 

When, |\arg {z_1} - \arg {z_2}| = \frac{\pi }{2}

 \arg \frac{{{z_1}}}{{{z_2}}} = \pm \frac{\pi }{2}

 i.e. \frac{{{z_1}}}{{{z_2}}} \Rightarrow  Purely Imaginary
so, Option (d) is also correct.

 3. If |{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2} then ,

 a) |\arg {z_1} - \arg {z_2}| = \frac{\pi }{2}
b) {z_1}\overline {{z_2}} + \overline {{z_1}} {z_2} = 0
c) \frac{{{z_1}}}{{{z_2}}} \Rightarrow  Purely Real
d) \frac{{{z_1}}}{{{z_2}}} \Rightarrow  Purely Imaginary

 Sol: Since, |{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2} + 2{{\rm Re}\nolimits} ({z_1}\overline {{z_2}} )  [for explanation click here]

  [Since {z_1}\overline {{z_2}} + {z_2}\overline {{z_1}} = 2{{\rm Re}\nolimits} ({z_1}\overline {{z_2}} )  (? click here) ]

therefore, {z_1}\overline {{z_2}} + \overline {{z_1}} {z_2} = 0 and option (b) is correct

since, |{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2}

and from vector addition, vector {z_1}, {z_2} and {z_1} + {z_2} must form right angle triangle (according to Pythagoras Theorem)



therefore, angle between {z_1} and {z_2} is pi/2
 => |\arg {z_1} - \arg {z_2}| = \frac{\pi }{2}

 therefore, option (a) is also correct.

 Further,

 since, |{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2}

 or, |{z_2}{|^2}{\left| {\frac{{{z_1}}}{{{z_2}}} + 1} \right|^2} = |{z_2}{|^2}\left( {{{\left| {\frac{{{z_1}}}{{{z_2}}}} \right|}^2} + 1} \right)

 or, {\left| {\frac{{{z_1}}}{{{z_2}}} + 1} \right|^2} = \left( {{{\left| {\frac{{{z_1}}}{{{z_2}}}} \right|}^2} + 1} \right)

 let, {\frac{{{z_1}}}{{{z_2}}} = w}

 therefore, |w + 1{|^2} = |w{|^2} + {1^2}

 => angle between w and 1 is pi/2
since 1 is real number and lies on real axis, w must be lie on imaginary axis

 i.e. |arg.w| = pi/2

 therefore, w must be Purely imaginary
or, \frac{{{z_1}}}{{{z_2}}} \Rightarrow  Purely Imaginary

 so, Option (d) is also correct.



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