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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def odt_to_ri(f, res, nm):
 r"""Convert the ODT object function to refractive index In :abbr:`ODT (Optical Diffraction Tomography)`, the object function is defined by the Helmholtz equation .. math:: f(\mathbf{r}) = k_\mathrm{m}^2 \left[ \left( \frac{n(\mathbf{r})}{n_\mathrm{m}} \right)^2 - 1 \right] with :math:`k_\mathrm{m} = \frac{2\pi n_\mathrm{m}}{\lambda}`. By inverting this equation, we obtain the refractive index :math:`n(\mathbf{r})`. .. math:: n(\mathbf{r}) = n_\mathrm{m} \sqrt{\frac{f(\mathbf{r})}{k_\mathrm{m}^2} + 1 } Parameters f: n-dimensional ndarray The reconstructed object function :math:`f(\mathbf{r})`. res: float The size of the vacuum wave length :math:`\lambda` in pixels. nm: float The refractive index of the medium :math:`n_\mathrm{m}` that surrounds the object in :math:`f(\mathbf{r})`. Returns ------- ri: n-dimensional ndarray The complex refractive index :math:`n(\mathbf{r})`. Notes ----- Because this function computes the root of a complex number, there are several solutions to the refractive index. Always the positive (real) root of the refractive index is used. """

    km = (2 * np.pi * nm) / res
    ri = nm * np.sqrt(f / km**2 + 1)
    # Always take the positive root as the refractive index.
    # Because f can be imaginary, numpy cannot return the correct
    # positive root of f. However, we know that *ri* must be postive and
    # thus we take the absolute value of ri.
    # This also is what happens in Slaneys
    # diffract/Src/back.c in line 414.
    negrootcoord = np.where(ri.real < 0)
    ri[negrootcoord] *= -1
    return ri