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Suppose $u(x,y),v(x,y)$ are real valued functions on an open set $A \subset R^2$. Suppose they satisfy cauchy riemann equations. Then

$$ u_1(x,y) = u^2 - v^2 , \; \; \; v_1(x,y) = 2uv $$

satisfy C-R as well. This follows from the chain rule:

$$ \frac{ \partial u_1}{\partial x} = 2u u_x - 2 v v_x = 2( u u_x - v v_x) $$ $$ \frac{ \partial u_1}{\partial y} = 2u u_y - 2 v v_y = 2( uu_y - vv_y )$$

$$ \frac{ \partial v_1}{\partial x} = 2u_xv + 2uv_x = 2(u_xv + uv_x) = 2(v_yv -uu_y) = - 2 (uu_y -v_y v) = - \frac{ \partial u_1}{\partial y}$$

Similarly, $(u_1)_x = (v_1)_y $. Hence, $u_1,v_1$ satisfy the C-R equations as well.

Question: Suppose we are given $u_2(x,y) = e^u \cos v $ and $v_2(x,y) = e^u \sin v $. IS there a way to make $u_2$ and $v_2$ in the form of $u_1$, $v_2$ so that we can have that they satisfy the cauchy riemman equations as well?

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  • $\begingroup$ To clarify: you want a proof that's not just "compositions of analytic functions are analytic"? $\endgroup$ – Chappers Sep 27 '15 at 0:45
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Let $a+ib:X \to Y$, and $u+iv:Z \to X$ be analytic, so we have enough continuity and $a(\cdot,\cdot)+ib(\cdot,\cdot)$ satisfies $$ \partial_1 a= \partial_2 b, \quad \partial_1 a = -\partial_1 b, \tag{1} $$ $u(x,y)+iv(x,y)$ satisfies $$ u_x = v_y, \quad u_y = -v_x. \tag{2} $$ Then $a(u,v)+ib(u,v):Z \to Y$ is analytic: $$ \partial_x a(u,v) = \partial_1 a u_x + \partial_2 a v_x = \partial_2 b \partial_y v + (-\partial_1 b)(-u_y) = \partial_y b(u,v), $$ using the chain rule and (1) and (2). The other equation works in the same way: $$ \partial_y a(u,v) = \partial_1 a u_y + \partial_2 a v_y = (\partial_2 b)(-v_x)+(-\partial_1 b)(u_x) = -\partial_x b(u,v). $$ This works for any such $a,b,u,v$, so the composition of real-valued Cauchy–Riemann-satisfying functions also satisfies C–R.

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