Subscript[k, 1][\[Epsilon]_] =
Sqrt[(\[Epsilon] - Subscript[v, 1])^2 -
Subscript[k, y][\[Epsilon]]^2]
目的是想要，当k1是实数的时候取k1，是虚数的时候取i*k1,以下是我的写法这么写对吗？
Subscript[\[CurlyPhi], 0] = 15*\[Pi]/180
Subscript[v, 1] = 1
Subscript[v, 2] = 1
Subscript[s, 0] = Sign[\[Epsilon]]
Subscript[s, 1] = Sign[\[Epsilon] - Subscript[v, 1]]
Subscript[s, 2] = Sign[\[Epsilon] - Subscript[v, 2]]
Subscript[k, 0][\[Epsilon]_] = \[Epsilon]*
Cos[Subscript[\[CurlyPhi], 0]]
Subscript[k, y][\[Epsilon]_] = \[Epsilon]*
Sin[Subscript[\[CurlyPhi], 0]]
Subscript[k, 11][\[Epsilon]_] =
Sqrt[(\[Epsilon] - Subscript[v, 1])^2 - Subscript[k, y][\[Epsilon]]^2]
Subscript[k, 12][\[Epsilon]_] =
i*(Sqrt[Subscript[k,
y][\[Epsilon]]^2 - (\[Epsilon] - Subscript[v, 1])^2])
Subscript[k, 1][\[Epsilon]_] =
Which[\[Epsilon] > (Subscript[v,
1]/(1 - Sin[Subscript[\[CurlyPhi], 1]])),
Subscript[k,
11][\[Epsilon]], \[Epsilon] < (Subscript[v,
1]/(1 + Sin[Subscript[\[CurlyPhi], 1]])),
Subscript[k,
11][\[Epsilon]], \[Epsilon] <= (Subscript[v,
1]/(1 - Sin[Subscript[\[CurlyPhi], 1]])),
Subscript[k,
12][\[Epsilon]], \[Epsilon] >= (Subscript[v,
1]/(1 + Sin[Subscript[\[CurlyPhi], 1]])),
Subscript[k, 12][\[Epsilon]]]
另外，想求知有没有什么更简便的方法。谢谢~

……你为什么不套个Abs呢？