For z= x+iy ,
cosz = 1/2 (exp(iz)+exp(-iz))
sinz = 1/2i (exp(iz)-exp(-iz))
Evaluate Re{ (1+i)sin {(2+i)(π/4)} }
Simplify the answer as much as possible
complex number question ( urgent ) PLEASE HELP?
2014-04-13 7:52 am
回答 (1)
2014-04-15 7:01 am
Re {(1+i) sin((2+i)(π/4))}
= Re {(1+i) sin(π/2 + πi/4)}
= Re {(1+i) (sin(π/2) cos(πi/4) + cos(π/2) sin(πi/4))}
= Re {(1+i) (1 cos(πi/4) + 0)}
= Re {(1+i) cos(πi/4)}
= Re {(1+i) * (1/2) (e^(i * πi/4) + e^(-i * πi/4))}
= Re {(1/2)(1 + i) * (e^(-π/4) + e^(π/4))}
= Re {[(e^(-π/4) + e^(π/4))/2 + i (e^(-π/4) + e^(π/4))/2]}
= (e^(-π/4) + e^(π/4))/2
= cosh(π/4).
I hope this helps!
= Re {(1+i) sin(π/2 + πi/4)}
= Re {(1+i) (sin(π/2) cos(πi/4) + cos(π/2) sin(πi/4))}
= Re {(1+i) (1 cos(πi/4) + 0)}
= Re {(1+i) cos(πi/4)}
= Re {(1+i) * (1/2) (e^(i * πi/4) + e^(-i * πi/4))}
= Re {(1/2)(1 + i) * (e^(-π/4) + e^(π/4))}
= Re {[(e^(-π/4) + e^(π/4))/2 + i (e^(-π/4) + e^(π/4))/2]}
= (e^(-π/4) + e^(π/4))/2
= cosh(π/4).
I hope this helps!
收錄日期: 2021-05-01 00:02:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140412235221AAXlFty