(z+i)^n+(z-i)^n=0
z member of C
n member of N
2006-10-22 19:58:53 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Can you give the procedure and not just the result.
2006-10-22 20:11:28 · update #1
devide by (z-i)^n
((z+i)/(z-i))^n + 1 =0
let (z+i)/(z-i) = t
t^n+ 1 = 0
this can be solved for n values of t
t^n = -1 = e^(pi*i) = e^(i*pi(2k+1)) for k from 0 to n- 1
this can be solved for t
then this can be solved for z
2006-10-22 20:13:01 · answer #1 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
Review your math book again and again until you understand better. Or go to your teacher for help.
2006-10-23 03:00:01 · answer #2 · answered by yahooaddict 4 · 0⤊ 1⤋
z = cos u + isin u
then,
(z+i)^n = [(cos u + isin u) + i]^n = cos nu + isin nu + P(u) + i^n
(z-i)^n = [(cos u + isin u) - i]^n = cos nu + isin nu - P(u) - i^n
adding,
2cos nu + 2isin nu = 0
cos nu + isin nu = 0
(cos u + isin u)^n = 0
e^iu = 0
2006-10-23 04:45:46 · answer #3 · answered by yasiru89 6 · 0⤊ 0⤋
It equals 2z^n
2006-10-23 03:03:16 · answer #4 · answered by Cycleogical 2 · 0⤊ 0⤋