QUIZ 5.5B

Lemma If f is analytic, and |f| is constant, then f is constant.

Proof Let f = u + iv. Then we are given that {  1  }.   So

{  2  },   2uuy + 2vvy = 0, leading to {  3  }.

Also, by the Cauchy-Riemann equations, ux = vy,  uy = -vx.

So, eliminating the u terms, we get vx2 + vy2 = 0  and so vx = 0 = vy = ux = uy.

Hence {  4  } and so f(z) is constant.

Match the above boxes 1, 2, 3, 4 with the selections
(a)  f '(z) = 0 ;   (b)  ux/uy = vx/vy ;  (c)  2uux + 2vvx = 0 ;  (d)  u2 + v2 = c .

My solutions:

1.   ;  2.   ;  3.   ;  4.   .   Check