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Solving Complex Variables Equations


Date: 09/01/98 at 08:12:41
From: derek zumpolle
Subject: Complex numbers: determining z

Determine all z such that:

   2z + (conjugate of z)^2 = -1+6i
   4z^2 - (conjugate of z)^2 = 20i
   (conjugate of z) z^3 = 4i


Date: 09/01/98 at 08:48:38
From: Doctor Jerry
Subject: Re: Complex numbers: determining z

Hi Derek,

If z = x+i*y, then the first equation is:

              2(x+i*y) + (x-i*y)^2 = -1+6i
   2x + 2*i*y + x^2 - y^2 - 2i*x*y = -1+6i

So:

                    2x + x^2 - y^2 = -1
                         2y - 2x*y = 6

Solve this system simultaneously for x and y. The others can be done 
in the same way.

- Doctor Jerry, The Math Forum
  Check out our web site! http://mathforum.org/dr.math/   


Date: 09/01/98 at 11:51:39
From: Doctor Anthony
Subject: Re: complex numbers: determining z

I shall use z* to represent the conjugate of z. 

To find the z such that 2z + (z*)^2 = -1+6i:

                        2z + (z*)^2 = -1+6i
      2(x+iy) + [x^2 - i 2xy - y^2] = -1 + 6i
       x^2 + 2x - y^2 + i[2y - 2xy] = -1 + 6i
                     x^2 + 2x - y^2 = -1   and   2y(1-x) = 6  

So:                               y = 3/(1-x)

Substituting for y:

               x^2 + 2x - 9/(1-x)^2 = -1
              (x^2 + 2x + 1)(1-x)^2 = 9
                    (x+1)^2 (x-1)^2 = 9
                        [x^2 - 1]^2 = 9    

Taking square roots:

                            x^2 - 1 = +/-3   

So x^2 = -2 (which is not possible as x is real)
or x^2 = 4.

Thus x = +/- 2.

Then if x = 2, y = 3/(1-2) = -3
 and if x = -2, y = 3/(1+2) = 1

So two solutions are z = 2 - 3i and z = -2 + i.

For 4z^2 - (z*)^2 = 20i, factorize the lefthand side:

   (2z-z*)(2z+z*) = 20i

Since:

   2z = 2x + 2iy
   z* =  x -  iy

we have:

   2z - z* = x + 3iy
   2z + z* = 3x + iy

Remultiplying the brackets:

   3x^2 + 10ixy - 3y^2 = 20i

Therefore:

   3x^2 - 3y^2 = 0    
           y^2 = x^2   
             y = +/-x

and:

          10xy = 20  
            xy = 2   

Thus:

   x = sqrt(2),  y = sqrt(2)    or 
   x = -sqrt(2), y = -sqrt(2)

Then the two solutions are z = sqrt(2)[1 + i] and z = -sqrt(2)[1 + i].

For (z*) z^3 = 4i:

                               (x - iy)^3 = 4i
   x^3 + 3x^2 (-iy) + 3x(-iy)^2 + (-iy)^3 = 4i
             x^3 - 3xy^2 + iy[y^2 - 3x^2] = 4i

So   x(x^2 - 3y^2) = 0  and y[y^2 - 3x^2] = 4

Thus x = 0 and y^3 = 4, which implies y = 4^(1/3), 
or x^2 = 3y^2, which implies:

   y[y^2-9y^2] = 4
         -8y^3 = 4
           y^3 = -1/2
             y = (-1/2)^(1/3)

Thus:

           x^2 = 3(-1/2)^(2/3)
             x = sqrt[3(-1/2)^(2/3)]

So the two solutions are:

             z = 0 + i 4^(1/3)
             z = sqrt[3(-1/2)^(2/3)] + i (-1/2)^(1/3)

- Doctor Anthony, The Math Forum
  Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Imaginary/Complex Numbers

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