先告訴你個(gè)公式：sin(a+bi)=[e^b+e^(-b)]/2*sina+i*[e^b-e^(-b)]/2*cosa
設(shè)z=a+bi,則z+i=a+(b+1)i
sin(z+i)=1
sin[a+(b+1)i]=1
[e^(b+1)+e^(-b-1)]/2*sina+i*[e^(b+1)-e^(-b-1)]/2*cosa=1
因此,①[e^(b+1)+e^(-b-1)]/2*sina=1 ②[e^(b+1)-e^(-b-1)]/2*cosa=0
由②得,a=π/2或a=-π/2或b=-1
當(dāng)a=π/2時(shí),e^(b+1)+e^(-b-1)=2
[e^(b+1)]^2-2e^(b+1)+1=0
[e^(b+1)-1]^2=0
b=-1
當(dāng)a=-π/2時(shí),e^(b+1)+e^(-b-1)=-2