es for order ``v=1`` at ``z=1000j``. We see that
    `jv` overflows but `jve` returns a finite number:

    >>> import numpy as np
    >>> from scipy.special import jv, jve
    >>> v = 1
    >>> z = 1000j
    >>> jv(v, z), jve(v, z)
    ((inf+infj), (7.721967686709077e-19+0.012610930256928629j))

    For real arguments for `z`, `jve` returns the same as `jv`.

    >>> v, z = 1, 1000
    >>> jv(v, z), jve(v, z)
    (0.004728311907089523, 0.004728311907089523)

    The function can be evaluated for several orders at the same time by
    providing a list or NumPy array for `v`:

    >>> jve([1, 3, 5], 1j)
    array([1.27304208e-17+2.07910415e-01j, -4.99352086e-19-8.15530777e-03j,
           6.11480940e-21+9.98657141e-05j])

    In the same way, the function can be evaluated at several points in one
    call by providing a list or NumPy array for `z`:

    >>> jve(1, np.array([1j, 2j, 3j]))
    array([1.27308412e-17+0.20791042j, 1.31814423e-17+0.21526929j,
           1.20521602e-17+0.19682671j])

    It is also possible to evaluate several orders at several points
    at the same time by providing arrays for `v` and `z` with
    compatible shapes for broadcasting. Compute `jve` for two different orders
    `v` and three points `z` resulting in a 2x3 array.

    >>> v = np.array([[1], [3]])
    >>> z = np.array([1j, 2j, 3j])
    >>> v.shape, z.shape
    ((2, 1), (3,))

    >>> jve(v, z)
    array([[1.27304208e-17+0.20791042j,  1.31810070e-17+0.21526929j,
            1.20517622e-17+0.19682671j],
           [-4.99352086e-19-0.00815531j, -1.76289571e-18-0.02879122j,
            -2.92578784e-18-0.04778332j]])
    Ú