More Complex Analysis

Let u(x,y) and v(x,y) be real-valued functions on an open set A ⊂ ℝ^2 = ℂ.
Suppose that they satisfy the Cauchy-Riemann equations on A.
Show that

u_1(x,y) = [u(x,y)]^2 - [v(x,y)]^2 and v_1(x,y) = 2u(x,y) v(x,y)

satisfy the Cauchy-Riemann equations on A and that the functions

u_2(x,y) = e^[u(x,y)] cos v(x,y) and v_2(x,y) = e^[u(x,y)] sin v(x,y)

also satisfy the Cauchy-Riemann equations on A.

Can you do this without performing any calculations?


Thanks!

I don't think so since it has to be shown that they satisfy a certain system of PDE.

Since u and v satisfy the Cauchy-Riemann equations on A, ∂u/∂x = ∂v/∂y and ∂u/∂y =
-∂v/∂x on A.

For the first part, we have that, on A,

∂(u_1)/∂x = 2u(∂u/∂x) - 2v(∂v/∂x) = 2u(∂v/∂y) - 2v(-∂u/∂y) = 2u(∂v/∂y) + 2v(∂u/∂y) = ∂(v_1)/∂y,

∂(u_1)/∂y = 2u(∂u/∂y) - 2v(∂v/∂y) = 2u(-∂v/dx) - 2v(-∂u/∂x) = -2u(∂v/∂x) + 2v(∂u/∂x) =
-∂(v_1)/∂x.

So u_1, v_1 satisfy the Cauchy - Riemann equations on A.

For the second part, we have that, on A,

∂(u_2)/∂x = e^u (-sin v • ∂v/∂x) + (e^u • ∂u/∂x) cos v

= -e^u sin v (-∂u/∂y) + ∂v/∂y e^u cos v

= (e^u • ∂u/∂y) sin v + e^u (cos v • ∂v/∂y)

= ∂/∂y(e^u) sin v + e^u • ∂/∂y(sin v)

= ∂/∂y(e^u sin v)

= ∂(v_2)/∂y,

and

∂(u_2)/∂y = e^u (-sin v • ∂v/∂y) + (e^u • ∂u/∂y) cos v

= -e^u (sin v • ∂u/∂x) + (e^u • -∂v/∂x) cos v

= -(e^u • ∂u/∂x)sin v - e^u (cos v • ∂v/∂x)

= -∂/∂x(e^u) sin v - e^u ∂/∂x(sin v)

= -∂/∂x(e^u sin v)

= -∂(v_2)/∂x.

Hence, the functions u_2, v_2 satisfy the Cauchy-Riemann equations are satisfied on A.