Laplacian

Needs["VectorAnalysis`"]

Laplacian of a harmonic function equals 0, whether it is in simple trigonometic form,
as in the example where the displacement that is harmonic

Needs["VectorAnalysis`"]
u={Cos[theta],b,c}
Laplacian[u]

or whether it is in a form that uses Euler's identity.  For example,  let's define a 
displacement that is complex harmonic.  By the Euler formula or Euler Identity:

		 (e^(iθ))== cosθ + i sinθ

We can make the exponent scalar,

Needs["VectorAnalysis`"]
Clear[u]
Laplacian[  ExpToTrig [Exp[I xX] ] ]

or, we can make the exponent vectorial

Needs["VectorAnalysis`"]
vfield ={0,0,Cos[theta]}
Laplacian[ ExpToTrig [Exp[I *vfield]  ] ]

In fact, we can use this to our advantage to test whether a function is harmonic.+

Let us now try to estimate a Laplacian of a vectorial function

Needs["VectorAnalysis`"]
v:={Xx^2+Zz, Yy*Sin[Xx], E^(3Zz)}
Laplacian[v]

a := Laplacian[Xx^2+Zz]
b := Laplacian[Yy*Sin[Xx]]
c := Laplacian[E^(3Zz)]

{a,b,c}

Let us now try to estimate a Laplacian of a scalar function

w := (Xx^2)+Zz+Yy Sin[Xx]+(E^(3 Zz))
d= Div[Grad[w]]

Laplacian[w]