Solutions to Complex Functions (2)

1)  Given u(x,y) + iv(x,y) = 2x² + i2y²   find f(z,z*)


Solution:

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

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