Lesson Upper level problems on complex numbers

Upper level problems on complex numbers

Problem 1

The equation in the post is 

   |      z                |
   | ----------  - conj(z) |  = 1 + |z|    (1)
   |  |conj(z)|            |


Vertical lines denote absolute value, or the modulus of complex numbers.


|conj(z)| is the same as |z|.

        z                          z
So, ----------   is the same as  ----,  which is the unit vector, co-directed with z.
     |conj(z)|                    |z|


So, equation (1) tells us that a unit vector, co-directed with z, MINUS conj(z) 
has the absolute value (the modulus) of 1+|z|.    (2)


Taking into account the rule of adding/subtracting complex numbers as vectors,
we can interpret the statement (2) THIS WAY :


    we have a triangle with three sides: side  "a"  is the unit vector, co-directed with z;

                                         side  "b"  is the vector -conj(z);

                                         side  "c"  has the length 1+|z|.


Keeping in mind a triangle inequality, we can conclude that this triangle is degenerated
and its sides "a", "b" and "c" all are collinear.  Namely,

    +-------------------------------------+
    |    -conj(z) is co-directed with z   |
    +-------------------------------------+


which may happen IF and ONLY IF  z is a purely imaginary complex number.


It is the statement to prove, and at this point, the solution is COMPLETE
and the problem' statement is proven.

Problem 2

In cis-form,  z = .   It means that the modulus of z is 1 and the argument of z is .

Then   = .  In other words,   has the modulus 1  and the argument of  .


To find the sum  z + ,  you should use the parallelogram rule of adding complex numbers as vectors.


Since both addends,  z  and    have equal modulus, the parallelogram rule in this case becomes
the rhombus rule.  In rhombus, the diagonal is the bisector of the angle.  
THEREFORE, in our case the argument of    is

     = . 


The modulus of    is the third side of an isosceles triangle with the equal legs of the length 1 
and the angle between these legs of   = 

    A =  = .


So,   = .



Similarly, for the denominator

     =  = ,


where  B = .   <<<---=== Note that the angle between  z  and    is  
                                              = 

                                             and   = .


Thus the ratio    has the modulus  

     =    


and the argument as the difference

     =  = .


Using formulas    = ,   = ,  the modulus of the ratio can be simplified


      =  = 



Therefore, the ratio    is  ,  or, which is the same, (,).


ANSWER.  In polar form,    is   (,).