→ playerOrz : 1. (z+1/z)(z^2-1+1/z^2) ? (z+1/z)^3-3(z+1/z) 11/26 14:52
→ angel07 : right 11/26 17:46
※ 編輯: angel07 (114.34.149.196), 11/26/2014 17:46:36
※ 引述《annie392 (安妮)》之銘言：
: (1)如果z+1/z是整數,則z^3+1/z^3 也是整數
: (2)如果z+1/z是有理數,則z也是有理數
: (3)如果z和1/z的實部相乘小於1,則z不是實數
: (4)如果z不是實數,但z+1/z是實數,則|z|=1
: (5)如果z和z+1/z的虛部都是正數,則|z|>1
: 可以詳細解說嗎?謝謝!
Sol:
Known that z is complex number, z!=0
z^2+1/z^2=(z+1/z)^2-2
1. Known that z+1/z is integer
Then z^3+1/z^3=(z+1/z)(z^2-1+1/z^2)=(z+1/z)((z+1/z)^2-3) is integer
=> True
2. Set z=i, then 1/z=-i
Therefore, z+1/z=0 is rational number, but z is NOT rational number
=> False
3. Set z=a+bi, then 1/z=1/(a+bi)=(a-bi)/(a^2+b^2)
If a．a/(a^2+b^2)=a^2/(a^2+b^2)<1, then b!=0, i.e. a is NOT real number
=> True
4. Set z=a+bi, b!=0, z+1/z=(a+bi)+(a-bi)/(a^2+b^2) is real number
That is, b-b/(a^2+b^2)=b(a^2+b^2-1)/(a^2+b^2)=0
Therefore, a^2+b^2-1=0 => |z|=sqrt(a^2+b^2)=1
=> True
5. Set z=a+bi, b>0 AND b(a^2+b^2-1)/(a^2+b^2)>0
Then a^2+b^2-1>0 => |z|=sqrt(a^2+b^2)>1
=> True
--
Monster_Reborn Heavy_Storm Dark_Hole Book_of_Moon Pot_of_Averice Foolish_Burial
Mystical_Space_Typhoon Forbidden_Lance Forbidden_Chalice Pot_of_Duality
Compulsory_Evacuation_Device Fiendish_Chain Breakthrough_Skill Effect_Veiler
Maxx_C D.D._Crow Torrential_Tribute Solemn_Judgment Solemn_Warning Mirror_Force
Bottemless_Trap_Hole Gorz_the_Emissary_of_Darkness Tragordia Dragon_Rulers
Soul_Charge Night_Beam Instant_Fusion Super_Polymerization Vanity's_Emptiness
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 140.115.189.46
※ 文章網址: http://www.ptt.cc/bbs/Math/M.1416979421.A.B8D.html