Let $$f\left( z \right) = \sin \,z$$   and $$g\left( z \right) = \cos \,z.$$   If $$*$$ denotes a composition of functions, then the value of $$\left( {f + ig} \right)*\left( {f - ig} \right)\left( z \right)$$    is:

Solution :
$$\eqalign{ & \left( {f - ig} \right)\left( z \right) \cr & = f\left( z \right) - ig\left( z \right) \cr & = \sin \,z - i\,\cos \,z \cr & = - i\left( {\cos \,z + i\,\sin \,z} \right) \cr & = - i{e^{iz}} = \theta \,\,\left( {{\text{say}}} \right) \cr & {\text{Now, }}\left( {f + ig} \right)*\left( {f - ig} \right)\left( z \right) \cr & = \left( {f + ig} \right)\left( {f - ig} \right)\left( z \right) \cr & = \left( {f + ig} \right)\left( \theta \right) \cr & = f\left( \theta \right) + ig\left( \theta \right) \cr & = \sin \,\theta + i\,\cos \,\theta \cr & = i\left( {\cos \,\theta - i\,\sin \,\theta } \right) \cr & = i{e^{ - i\theta }} \cr & = i{e^{ - i\left( { - i{e^{iz}}} \right)}} \cr & = i{e^{ - {e^{iz}}}} \cr} $$