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由 mather5012 於 星期四 三月 24, 2005 8:11 pm
(z-5)/(z'-5)=(z-5)/(z-5)',令(z-5)=r(cosA+isinA),r>0,則(z-5)'=r(cosA-isinA),
(z-5)/(z'-5)=(cosA+isinA)/(cosA-isinA)=cos(2A)+isin(2A),故a=cos(2a),z=
(rcosA+5)+irsinA,|z|^2=9=(rcosA+5)^2+(rsinA)^2=r^2+10rcosA+16=0,
cosA=[(-r^2-16)/(10r)]-1,a=cos(2A)=2(cosA)^2-1=2[(r^2+16)/(10r)]^2-1
=(1/50)(r+(16)/r)^2-1>=(1/50)[2(root(r(16/r))]^2-1=64/50-1=7/25
Re: [數學]求最小值?
由 雞腿飯 於 星期二 三月 15, 2005 10:42 pm
求a的最小值?
|z|=3 , (z-5)/(z' -5)=a+bi
let m=z-5
m/m'=cos2t+isin2t
t=angle(z-5)=asin0.6~-asin0.6
2t=2asin0.6~-2asin0.6
cos2t=1~cos(2asin0.6)
=1~cos(asin0.6)^2-0.6^2
=1~0.28
min=0.28
|z|=3 , (z-5)/(z' -5)=a+bi
let m=z-5
m/m'=cos2t+isin2t
t=angle(z-5)=asin0.6~-asin0.6
2t=2asin0.6~-2asin0.6
cos2t=1~cos(2asin0.6)
=1~cos(asin0.6)^2-0.6^2
=1~0.28
min=0.28
由 Searchtruth 於 星期五 八月 08, 2003 6:43 pm
好複雜喔..
我公佈答案...7/25
我公佈答案...7/25
由 bugzpodder 於 星期四 八月 07, 2003 11:11 pm
let z=3cis(t)
(z-5)/(z'-5)
=(z-5)^2/(z'z-5(z+z')+25)
=(z-5)^2/(28-30cos(t))
=(z^2-10z+25)/(28-30cos(t))
Re((z-5)/(z'-5))=(9cos(2t)-30cos(t)+25)/(28-30cos(t))
so we want the minimum of:(9(2cos^2(t)-1)-30cos(t)+25)/(28-30cos(t))
=18cos^2(t)-30cost+16/(28-30cost)
=(9cos^2(t)-15cost+8)/(14-15cost)
a little cleaner, what i did was transform (z-5)/(z'-5)=(z-5)^2/|z-5|^2
hmm use calculus? maybe there is an ingenious approach
(z-5)/(z'-5)
=(z-5)^2/(z'z-5(z+z')+25)
=(z-5)^2/(28-30cos(t))
=(z^2-10z+25)/(28-30cos(t))
Re((z-5)/(z'-5))=(9cos(2t)-30cos(t)+25)/(28-30cos(t))
so we want the minimum of:(9(2cos^2(t)-1)-30cos(t)+25)/(28-30cos(t))
=18cos^2(t)-30cost+16/(28-30cost)
=(9cos^2(t)-15cost+8)/(14-15cost)
a little cleaner, what i did was transform (z-5)/(z'-5)=(z-5)^2/|z-5|^2
hmm use calculus? maybe there is an ingenious approach
[數學]求最小值?
由 Searchtruth 於 星期三 八月 06, 2003 7:26 pm
已知 z屬於C , |z|=3 , (z-5)/(z' -5)=a+bi , a,b屬於R
求a的最小值?
ps z' 就是z的共軛複數...
求a的最小值?
ps z' 就是z的共軛複數...