[數學]複變函數...

[數學]複變函數...

(以高中幾年級或是大學幾年級之類ㄉ回答..)

Searchtruth

scsnake

Searchtruth

scsnake

Searchtruth

Searchtruth 寫到:複變函數學~那是一年級就會上的嗎?..

yll

Searchtruth

sinx=2不是無解嗎?

hycheah2000

X is imagative?
pls share wt is 複變函數

alexman442003

kaseikami

sinx=(e^ix-e^-ix)/2i
sinx=2   -->e^ix-e^-ix=4i
e^2ix-4ie^ix-1=0
e^ix=2i+sqrt[-4+4]=2i
-
ix=ln(2i)=ln2+i(Pi/2+2kPi),k是整數
x=pi/2+2kpi-iln2約為1.5708 - 1.31696 i+2kpi

Anonymous 寫到:sinx=(e^ix-e^-ix)/2i
sinx=2   -->e^ix-e^-ix=4i
e^2ix-4ie^ix-1=0
e^ix=2i+sqrt[-4+4]=2i
-
ix=ln(2i)=ln2+i(Pi/2+2kPi),k是整數
x=pi/2+2kpi-iln2約為1.5708 - 1.31696 i+2kpi

Anonymous 寫到:sinx=(e^ix-e^-ix)/2i
sinx=2   -->e^ix-e^-ix=4i
e^2ix-4ie^ix-1=0
e^ix=2i+sqrt[-4+4]=2i
-
ix=ln(2i)=ln2+i(Pi/2+2kPi),k是整數
x=pi/2+2kpi-iln2約為1.5708 - 1.31696 i+2kpi

sinx=2
(e^ix-e^-ix)/2i=2
e^ix-e^-ix=4i
e^2ix-4ie^ix-1=0
e^ix={4i+sqrt[-16+4]}/2 or {4i-sqrt[-16+4]}/2
e^ix=[2+sqrt(3)]i or [2-sqrt(12)]i
ix=ln{[2+sqrt(3)]i} or ln{[2-sqrt(12)]i}
ix=ln[2+sqrt(3)]+[2k(pi)+pi/2]i or ln[2+sqrt(12)]-[2k(pi)+pi/2]i
x=2k(pi)+pi/2-i ln[2+sqrt(3)] or 2k(pi)+pi/2-i ln[2-sqrt(3)]