已知f(z+i)=3z-2i,则f(i)=
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已知f(z+i)=3z-2i,则f(i)=
已知f(z+i)=3z-2i,则f(i)=
已知f(z+i)=3z-2i,则f(i)=
令Z=0,得f(i)= -2i.
若是求f(x),可如此:f(z+i)=3z-2i=3(z+i)-5i,故f(x)=3x-5i,进而f(i)=3i-5i=-2i.
已知f(z+i)=3z-2i,则f(i)=
已知f(z+i)=z+2z-2i,则f(i)=?
已知f(z)=2z+z^2+(1+i),则f(i)的值是
已知f(z)=2z+z²+(1+i)则f(i)的值是
f(z)=2z+z'-3i f(z'+i)=6-3i,则f(-z)=?z',z是复数!
f(z)=z^2/{(z^2+1)*(z^2+9)}求Res(z=i)f(z)和Res(z=3i)f(z)
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f(z+i)=3z-2i.则f(i)=还有为什么要等于零
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