Solutions To Complex Functions & Equations Revisited

The Problems again:

1)  (a) Given u(x,y) + iv(x,y) = 2x² + i2y²  

find f(z,z*)

(b)  Express f(z)= z² + z – 3 in polar form

2) Solve for x and y:   8x 2 + 3iy  -  4    = 8y   – 4iy

3) Solve: (z+1) 3  = z 3

Solutions:

(1)   Let:  z = x + iy, and z* = x – iy

Adding:   z + z* = (x + iy) + (x - iy) = 2x

We see: x = (z + z*)/2

Subtracting:

(z – z*) = [x + iy – x + iy] =  i2y

Then:  y = (z – z*)/ 2i

We can now formulate the function f(z,z*):

f(z,z*) = 2[(z + z*)/2]²  + i2[(z – z*)/ 2i]²

2)  z² = r² exp(i2(q)) = r² (cos (2q) + isin(2 q))

z = r exp(i(q)) = r(cos(q) + isin(q)

so: z² + z = r² (cos (2q) + isin(2 q)) +  r(cos(q) + isin(q)

Collecting like terms in i and simplifying:

f(z) =  r²(cos (2q) + r(cos(q)) + i{sin(2q) + sin(q)} – 3

so: iv(r, q) =  i{sin(2q) + sin(q)}

and v (r, q) =  {sin(2q) + sin(q)}

while:

u(r, q) =  r² (cos (2q) + r(cos(q)) – 3

3)     8x 2 + 3iy  -  4    = 8y   – 4iy

 8x 2    =   4

So:   x 2    =   4/ 8   =  1/ 2

Therefore:  x =  Ö  ( 1/2)  =   1/ Ö2

=  Ö 2  /  2

  

And:   

8 y   =   - 3iy  - 4iy   =   - 7 iy

y  =   -7 i/  8

4)  Expand the left side and set equal to the right:

z ³ + 3z ² +  3z + 1   =  z ³

è3z ² +3z + 1   = z ³    -    z ³
  
 or  3z ²  + 3z +1 =  0

(This can be solved using the quadratic formula, to give two roots)

Then:  z1 = ½ + i Ö (3) / 6 

And z2 = -(½ ) + iÖ (3)/ 6

Checking the result against the  original equation:

z ³  = 0.192i and  (z ³ + 1) ³  = 0.192i

So both quantities are equal, the roots are correct.