复数z=1+根号3 则z的600次方=

原理:
A=x+y*i=e^(a+α*i)=e^a*e^(α*i)=Ra*e^(α*i)=Ra*[cos(α)+sin(α)*i]
B=m+n*i=e^(b+β*i)=e^b*e^(β*i)=Rb*e^(β*i)=Rb*[cos(β)+sin(β)*i]
|A|=Ra=e^a=(x^2+y^2)^0.5,α=Arctan(y/x),明显a=Ln(|A|)
|B|=Rb=e^b=(m^2+n^2)^0.5,β=Arctan(n/m),明显b=ln(|B|)
A^B=[e^(a+α*i)]^(m+n*i)
=e^[(a+α*i)*(m+n*i)]
=e^[(a*m-α*n)+(a*n+α*m)*i]
=[e^(a*m-α*n)]*[e^((a*n+α*m)*i)]
=[e^(a*m-α*n)]*[cos(a*n+α*m)+sin(a*n+α*m)*i]
=[cos(a*n+α*m)*e^(a*m-α*n)]+[sin(a*n+α*m)*e^(a*m-α*n)]*i
解题:
A=z=1+3^0.5*i,则a=Ln[(1+3)^0.5]=Ln2,α=π/3
B=600,则m=600,n=0
z^600
=A^B
=[cos(a*n+α*m)*e^(a*m-α*n)]+[sin(a*n+α*m)*e^(a*m-α*n)]*i
=[cos(ln2*0+π/3*600)*e^(ln2*600-π/3*0)]+[sin(ln2*0+π/3*600)*e^(ln2*600-π/3*0)]*i
=[cos(200π)*e^(600*ln2)]+[sin(200π)*e^(600*ln2)]*i
=(e^ln2)^600
=2^600
=4149515568880992958512407863691161151012446232242436899995657329690652811412908146399707048947103794288197886611300789182395151075411775307886874834113963687061181803401509523685376

不知道

是z=1+i√3=2(cosπ/3+isinπ/3)
所以z³=2³*(-1)
则z^6=2^6
所以z^600=2^600