I list up this method to transform a complex function to Cartestian form, which can be used on virtually any function:

such as:

```
u0(r_, phi_) := Sum(I^(-n) BesselJ(n, r) Exp(I n phi), {n, -5, 5});
TransformedField("Polar" -> "Cartesian", u0(r, phi), {r, phi} -> {x, y})
```

which yields:

```
u0(x_, y_) :=
BesselJ(0, Sqrt(x^2 + y^2)) +
1/(x^2 + y^2)^(5/2)
2 (-I x (x^2 + y^2)^2 BesselJ(1, Sqrt(x^2 + y^2)) +
Sqrt(x^2 + y^2) (-x^4 + y^4) BesselJ(2, Sqrt(x^2 + y^2)) +
I x^5 BesselJ(3, Sqrt(x^2 + y^2)) -
2 I x^3 y^2 BesselJ(3, Sqrt(x^2 + y^2)) -
3 I x y^4 BesselJ(3, Sqrt(x^2 + y^2)) +
x^4 Sqrt(x^2 + y^2) BesselJ(4, Sqrt(x^2 + y^2)) -
6 x^2 y^2 Sqrt(x^2 + y^2) BesselJ(4, Sqrt(x^2 + y^2)) +
y^4 Sqrt(x^2 + y^2) BesselJ(4, Sqrt(x^2 + y^2)) -
I x^5 BesselJ(5, Sqrt(x^2 + y^2)) +
10 I x^3 y^2 BesselJ(5, Sqrt(x^2 + y^2)) -
5 I x y^4 BesselJ(5, Sqrt(x^2 + y^2)))
```

or

a Hankel and Bessel function together:

```
u(r_, phi_) := Piecewise({{BesselJ(1.5 r, 5)*Exp(I 5 phi),
0 < r < 1/2}, {(BesselJ(3 r, 5) + BesselY(3 r, 5))*Exp(I 5 phi),
1/2 < r < 1}, {HankelH1(r, 5)*Exp(I 5 phi), r > 1}})
```

which yields:

which yields the respective given plots, when plotted:

and

However, my supervisor thinks these plots look “strange”.

How can I verify that the TransformedField command did the right job – for such extended functions?

In other words, how to trobleshoot TransformedField?