Lesson Solving non-standard equations in complex numbers

Solving non-standard equations in complex numbers

Problem 1

It is better and much easier to analyse and to solve this problem in polar trigonometric form.


If z^2 = z complex conjugate,  then, firstly, the modulus of z is equal to 1.

In other words, z lies on the unit circle in a coordinate plane.


Next, if the argument of z is polar angle ,  then the polar angle of z^2 is  ,

while the polar angle of (z conjugate) is    or  .


So, we get the equation for the polar angle


    Case 1.   = -,  which implies   = 0,  or   = 0.   Then the solution is z = 1.


OR


    Case 2.   = ,  which implies   = .    Hence,   = .


ANSWER.  There are TWO solutions.  One solution is z = 1.

          The other solution is  z =  = .

Problem 2

Our starting equation is

     -  = .    (1)



Since  = -1,  this equation is the same as

     +  = .



We can write it in this equivalent form

     =  - .



We can cancel common factor " i " in both sides

     =  - .       (2)



Left side is a real number --- hence, right side must be a real number, too.



Ok.  Now let z = a+bi. Then from (2),  right side of (2) is

      -  = 3*(a^2 - 2abi - b^2) - (a^2 +2abi - b^2) =

   = (3a^2 - 3b^2- a^2 + b^2) + (-6abi - 2abi) = (2a^2 - 2b^2) - 8abi.    (3)


Since the right side of (2) is a real number, it implies from (3) that  ab = 0, i.e.

        EITHER a= 0  OR  b= 0   (or BOTH).



So, below we analyze two cases.


    (a)  b= 0. It means that "z" is, actually, a real number.

         It is easy to check, that then equation (2) is valid for any value of real number z,

         which means that any real number is the solution for the original equation.



    (b)  a= 0, b=/= 0.  It means that "z" is, actually, a pure imaginary number.

         Then in (2), the left side is positive real number, while the right side (3) 
         is negative real number, which creates a CONTRADICTORY.


        +-----------------------------------------------------------------------------+
        |         So, the    to the problem is THIS:                         |
        |    the solution set to the given equation is the set of all real numbers.   |
        +-----------------------------------------------------------------------------+