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Leaky Nun
Leaky Nun
8:39 AM
f(x+yi)=u(x,y)+iv(x,y)
u_x + i v_y = (1/i)(u_y + iv_y)
u_x = v_y and u_y + v_x = 0
now let |f| = constant, so |f|^2 = u^2 + v^2 = constant
2uu_x + 2vv_x = 0
2uu_y + 2vv_y = 0
2uv_y + 2vv_x = 0
-2uv_x + 2vv_y = 0
 
Leaky Nun
Leaky Nun
9:06 AM
f(z+h) = f(z) + f'(z) g(h)
|f(z+h)|^2 - |f(z)|^2 = f'(z) conj(g(h)) + conj(f'(z)) g(h) = 0
ok let's say f(0) = 1
uv_x = vv_y
(1-v^2) v_x^2 = v^2 v_y^2
if f(0)=1 then u_x = v_x = 0 right
f(0) = 1, i.e. u(0,0)=1 and v(0,0)=0
2uu_x + 2vv_x = 0, so u_x = 0
-2uv_x + 2vv_y = 0 so v_x = 0
by CR equations we also have u_y = v_x = 0
 
Leaky Nun
Leaky Nun
9:46 AM
now more generally:
uu_x + vv_x = 0
uu_y + vv_y = 0
uv_y + vv_x = 0
-uv_x + vv_y = 0
 

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