z=x+iy求Re(e^(1/z))

Question

z=x+iy 求Re(e^(1/z))

Answer 1

1/z=1/(x+iy)=1/(x²＋y²) (x-iy)
e^(1/z)=e^[1/(x²＋y²) (x-iy)]
=e^[x/(x²＋y²)]·e^(-y/(x²＋y²)i)
=e^[x/(x²＋y²)]{cos[-y/(x²＋y²)]+isin[-y/(x²＋y²)]}
所以
Re(e^(1/z))
=e^[x/(x²＋y²)]cos[-y/(x²＋y²)]