Question

How to rebuild the analytical function f(z) if given `U(x,y)=Re f(z)=2x+(x/(x^(2)+y^(2)))` `0<|z|<+oo`?

Answer 1

`f(z) = (2x+x/(x^2+y^2)) + 1i`

Explanation:

By the very definition of `|z|` we have `0 le |z| lt oo`, so the only restriction is to ensure that `|z| != 0`

Suppose that we define:

`f(z) = U(x,y) + V(x,y)i`
`\ \ \ \ \ \ \ = (2x+x/(x^2+y^2)) + V(x,y)i`

Where `V(x,y)` is an arbitrary real function

Then we require:

`|U^2(x,y)+V^2(x,y)| > 0`
`:. |(2x+x/(x^2+y^2)) + V(x,y)i| > 0`
`:. sqrt((2x+x/(x^2+y^2))^2 + V^2) > 0`
`:. (2x+x/(x^2+y^2))^2 + V^2 > 0`

Clearly:

`0 le (2x+x/(x^2+y^2))^2 < oo`

So we can choose any function `V(x,y) > 0`

Thus one such function is:

`f(z) = (2x+x/(x^2+y^2)) + 1i`

Answer 2

See below.

Explanation:

If `f(z)=u(x,y)+iv(x,y)` is analytic then

`(partial u)/(partial x)=(partial v)/(partial y)`
`(partial v)/(partial x) = -(partial u)/(partial y)`

These equations occurred already in the 18th century in J.L. d'Alembert's and L. Euler's studies on functions of a complex variable.

Here `u(x,y)=2x+x/(x^(2)+y^(2))` and

`(partial u)/(partial x) = 2 - (x^2 - y^2)/(x^2 + y^2)^2 = (partial v)/(partial y)`
`(partial u)/(partial y) =-(2 x y)/(x^2 + y^2)^2=-(partial v)/(partial x)`

but

`v_1(x,y)=int_0^x(2 x y)/(x^2 + y^2)^2dx+xi(y) = y/(x^2 + y^2)+xi(y)`

and also

`v_2(x,y)=int_0^y (2 - (x^2 - y^2)/(x^2 + y^2)^2)dy+zeta(x)=2 y - y/(x^2 + y^2)+zeta(x)`

but `v_1(x,y)-v_2(x,y) = 2y-xi(y)+eta(x)=0`

then `eta(x)=0` and `xi(y)=2y`

so finally

`v(x)=2y+ y/(x^2 + y^2)`

Concluding,

`f(z)=2x+x/(x^(2)+y^(2))+i(2y+ y/(x^2 + y^2))`