Question

求大神帮忙解一下这复变函数题

Answer 1

(1)(√3+i)^4
根据z^n=ρ^n(cos(nθ)+sin(nθ))
ρ^n=√(（√3）^2+1^2)^4=16
nθ=4arctan1/√3=2π/3
(√3+i)^4=16（cos(2π/3)+sin(2π/3)）=-8+i8√3

(2)(1+i)^(1/4)
根据z^(1/n)=ρ^(1/n)(cos(θ/n)+sin(θ/4))
ρ^(1/n)=(√2)^(1/4)=2^(1/16)
θ/n =[arctan(1/1)]/4=π/16
(1+i)^(1/4)=2^(1/16)cos(π/16)+i2^(1/16)sin(π/16)

(3)e^(3+i4)=e^3*e^(i4)=e^3cos(arctan4)+ie^3sin(arctan4)

(4)ln(3+i4)=a+ib
3+i4 =e^(a+ib)=e^a*e^(ib)
e^a=sqrt(3^2+4^2)=5
a=ln5
b=Arctan(4/3)

=>ln(3+i4)=ln5+iArctan(4/3)

(5)sin(5i)
根据sinz=1/(i2)(e^(iz)-e^(-iz))展开
sin(5i) =1/（i2）(e^(-5)-e^(5))
=-i/2(e^(-5)-e^(5))

Answer 2

cosθ+isinθ= e^(iθ)