1+i=e^z,求z
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1+i=e^z,求z
1+i=e^z,求z
1+i=e^z,求z
∵1+i=√2(cosπ/4+isinπ/4)=√2e^(iπ/4)=e^z
∴z=ln[√2e^(iππ/4)]=ln(√2)+lne^(iπ/4)=ln(√2)+iπ/4.
应该是√2e^z=√2(cosπ/4+isinπ/4)
所以z=iπ/4
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