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如何在复数域内求方程z^4+a=0(a>0)和e^(z+1) +2=0的解?
人气:288 ℃ 时间:2019-09-09 18:01:43
解答
z^4+a=0z^4=-az^4=ae^(iπ)则z=a^(1/4) e^(iπ+2ikπ)/4,k=0,1,2,3即z0=a^(1/4)e^(iπ/4)z1=a^(1/4)e^(i3π/4)z2=a^(1/4)e^(i5π/4)z3=a^(1/4)e^(i7π/4)e^(z+1)+2=0e^(z+1)=-2e^(z+1)=2e^(iπ)e^(z+1)=e^(ln2+iπ+i...
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