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
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
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