^{2}+ i2y

^{2 }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}+ z – 3 in polar form

z

^{2}= r^{2}exp(i2(q)) = r^{2}(cos (2q) + isin(2 q))
z = r exp(i(q)) = r(cos(q) + isin(q)

so: z

^{2 }+ z = r^{2}(cos (2q) + isin(2 q)) + r(cos(q) + isin(q)
Collecting like terms in i and simplifying:

f(z) = r

^{2}(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) = r

^{2}(cos (2q) + r(cos(q)) – 3

## No comments:

Post a Comment