【Ｚ会】数学-複素数平面１：ポイント整理の学習

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1 章複素数平面 1: ポイント整理 PMYA1-Z1J1-01 学習時間のめやす 45 分 1. 複素数平面 < 1-1 複素数平面 > a = a+ bi a b xy ^a bh xy 1 ^a bh a = a+ bi 複素数平面 x 実軸 y 虚軸 a A A ^ a h 点 a O < 1 - 共役な複素数 > a = a+ bi a b a-bi a 共役な複素数 a a a = ^a+ bih^a- bih = a - ^bih = a + b a a a b a a a+ b = a + b a- b = a - b ab = a b c m = ^ a h = a b b a, a = a a 0, a = -a a = a+ bi a = a-bi a a xy ^a bh ^a -b h - a = - ^a+ bih = -a-bi - a = -^a- bih = - a+ bi a -a a - a a a a a a -a a - a 共役な複素数の対称性 6

2 PMYA1-Z1J1-0 < 1 - 複素数の絶対値 > a O a 絶対値 a a = a+ bi a b 章複 a = a+ bi = a + b 7 a a = 0, a + b = 0, a = b = 0, a = 0 b = 0 a a a = a = a 複素数の絶対値 a b a = a a a = - a = a ab = a b a a = b b ^b ] 0h a = a+ bi a b a = a-bi a a = ^a+ bih^a- bih = a - ^bih = a + b = a a = a+ bi a b a = a- bi = a + ^- bh = a + b = a - a = -a- bi = ^- ah + ^- bh = a + b = a ab = ab$ ab = ab$ a b = ^a a h^b b h = a b = a b a a a a b = $ b = a ` = b b b b < 1-4 極形式 > 0 a = a+ bi a b P OP OP x i = a a = cos i b = si i a a = ^cos i+ i si ih a 極形式 a i a 偏角 a ag a ag agumet i 0 E i 1-1 i E i OP a 1 i ag a = i+ i 0 E i 1-1 i E i 1 a = 0 = a = 0 素数平面1:ポイント整理

3 PMYA1-Z1J1-0 複素数の極形式 a = a+ bi ^a ] 0h a = ^cos i+ i si ih = a = a + b i = ag a < - 1. 複素数の演算の図形的意味実数倍加法減法 > k a = a+ bi a b 0 ka = k^a+ bih = ka+ kbi P ^ka h O A ^ a h k 0 O A k 1 0 O A P O ka = k a b a ] 0 0 a b, b = ka k a 原点を含む点が一直線上に存在する条件 a = a+ bi b = c+ di a b c d a + b = ^a+ bih + ^c+ dih = ^a+ ch + ^b+ dhi b c d 複素数 b だけの平行移動 O A ^ a h B ^ b h P ^a+ bh OA OB 4 a b a+ b a b つの複素数の和と平行移動 0 a b 4 0 a a+ b b 4 8

4 PMYA1-Z1J1-04 a b a-b = a+ ^-bh O A ^ a h B ^ b h -b 章 Bl P ^a-bh OA OBl 4 複 B ^ b h Bl^-b h O 4 O P ^a-bh A ^ a h B ^ b h 4 4 A ^ a h B ^ b h O P ^a-bh a-b 複素数平面上の点間の距離 a b a-b < - 乗法除法 > a = a+ bi b = c+ di a b c d ab = ^a+ bih^c+ dih = ^ac- bdh + ^ad+ bchi a = 1^cos i1+ i si i1h b = ^cos i + i si i h ab = 1^cos i1+ i si i1h$ ^cos i + i si i h = 1 ^cos i1cos i - si i1 si i h + i1 ^cos i1si i + si i1 cos i h cos ^i1+ i h = cos i1cos i -si i1si i si ^i1+ i h = si i1cos i + cos i1si i ab ab = 1 " cos ^i1+ i h + i si ^i1+ i h, ab a O i O b = ^cos i + i si i h a 1 a a = 1 a 9 素数平面1:ポイント整理

5 PMYA1-Z1J1-05 i = cos + i si a = ^ cos i+ i si ih ia = i$ ^cos i+ i si ih = ' cos ci + m + i si ci+ m1 i a つの複素数の積 a b 0 a = ^cos i + i si i h b = ^cos i + i si i h ab = " cos ^i + i h + i si ^i + i h, ab = = a b ag ^abh = i + i = ag a+ ag b 1 1 a b ab a O ag b O b a = a+ bi b = c+ di b ] 0 a a+ bi ac+ bd bc-ad = = + i b c + di c + d c + d a = ^cos i + i si i h b = ^cos i + i si i h a 1^cos i1+ i si i1h$ ^cos i -i si i h = b ^cos i + i si i h$ ^cos i -i si i h 1 ^cos i1cos i + si i1 si i h 1 ^si i1 cos i -cos i1si i h = + i cos ^i - i h = cos i cos i + si i si i si ^i - i h = si i cos i -cos i si i a 1 1 = cos ^i1-i h + i si ^i1-i h b 1 = " cos ^i1- i h + i si ^i1-i h, a O -i O 1 0

6 PMYA1-Z1J1-06 つの複素数の商 a b 0 a = 1^cos i1+ i si i1h b = ^cos i + i si i h 章複a 1 = " cos ^i1- i h + i si ^i1-i h, b 1 a 1 a = = b b a ag = i1- i = ag a-ag b b a a b a O -ag b b O 1 b 1 a = 1 + i O z1 w = cos + i si = i z 1 z1= aw = c 1 + imi 1 = - + i a = 1 + i b = - + i z b O a a - b = c 1 + im - c- + im = -i O ^a - bhw = ^ -ihi = 1+ i O b z z = ^a- bhw+ b = ^1+ ih + c- + im = i 素数平面1:ポイント整理

7 PMYA1-Z1J1-07 原点以外の点における回転 a b i z 1 z1= ^cos i+ i si ih^a- bh + b a b i b z z = ^cos i+ i si ih^a- bh + b < - 1. ドモアブルの定理ドモアブルの定理 > 1 i z = cos i+ i si i z z = ^cos i+ i si ih = cos ^i + ih + i si ^i + ih = cos i + i si i z = ^cos i + i si ih^cos i+ i si ih = cos ^i + ih + i si ^i + ih = cos i + i si i z = cos i + i si i = 1 = k z k+1 = k z z = ^cos ki + i si kih^cos i+ i si ih = cos ^k+ 1hi + i si ^k+ 1hi = k+ 1 z z 0-1 = 1 z = z = = = cos ^- ih + i si ^-ih z cos i + i si i ^cos i+ i si ih = cos i + i si i ドモアブルの定理

8 PMYA1-Z1J1-08 < - 乗根 > z = 1 章 z 1 の乗根複 1 z = 1 z z = z z 1 z = cos i+ i si i z = ^cos i+ i si ih = cos i + i si i z = 1 cos i = 1 ) si i = 0 i = k ` i = k k 4 0 E i 1 i = z = cos 0+ i si 0 cos + i si cos + i si 1 i 1 i = i = - + ~ ~ -1- i = ~ 1 1 ~ ~ 素数平面1:ポイント整理

9 PMYA1-Z1J z = 1 z 1 z = cos i+ i si i z = ^cos i+ i si ih = cos i + i si i z = 1 cos i = 1 ) si i = 0 i = k k ` i = k 0 E i 1 i k = 0 1 g の乗根 z = 1 z k k z k = cos + i si ^k = 0 1 g -1h 4

10 複素数平面 1: 重要例題 PMYA1-Z1J1-10 学習時間のめやす各 10 分ポイント整理 1 章 1 a = + i 複a a+ a a- a ia i b = 1+ i b b i 0 E i i 1- i 1 着眼 b b b = b 1 1 ia = i+ i 解答 = - + i a ia - + i = i i = - 1 = -i a+ a i i = ^ + h + ^ - h = a- a ^ i i = + h - ^ - h = i 右上図答 b = ^1+ ih = 1+ $ i+ ^ih = - + 4i 答 b = b = ^- h + 4 = 5 答 b = 1 + = b = ^ 5 h = i = cos + i si 答 i 1 = c - im 1- i = 5 5 = ccos + i si m 答 5 素数平面1:重要例題

11 PMYA1-Z1J1-11 ポイント整理 1 1 a b a+ b + a- b = ^ a + b a b c a = b = c = = a+ b+ c a b c a + b + c ] 0 h ab + bc + ca a+ b + c 着眼 1 a = a+ bi a b a = a a a = b = c = 1 解答 1 a+ b + a-b = ^a+ bh ^a+ bh + ^a-bh ^a-bh = ^a+ bh^ a+ b h + ^a-bh^ a- b h = ^a a+ a b + a b + b b h + ^a a-a b - a b + b b h = a a+ b b = ^ a + b h a = b = c = 1 a a = 1 b b = 1 c c = 1 ` 1 = a 1 = b 1 = c a b c = a+ b + c a b c = a+ b+ c = a+ b+ c ab + bc + ca = a+ b + c abc ` a = b = c = 1 ab + bc + ca = abc a+ b + c ab + bc + ca = a b c = 1 答 a+ b + c a = a a a! b = a! b a = a a = 1 z w = z w 6

12 PMYA1-Z1J1-1 ポイント整理 - 1 a = - 1+ i b = 1+ i c = + ai a a 章複 z w w ] 0 着眼 z = kw k a b c a b c a b c -a a 解答 O a b c, 0 b -a c -a, c - a = k^b -ah k c - a = ^+ aih - ^- 1+ ih = 4+ ^a-1hi b - a = ^1+ ih - ^- 1+ ih = + i c - a 4+ ^a-1hi = k ^b - ah = k+ ki b -a + i c - a = k^b -ah " 4+ ^a-1hi, ^1-ih = ^1+ ih^1-ih 4 = k a a 5 i = ^ + h + ^ - h a- 1 = k 4 a = 5 a- 5 = 0 ` a = 5 答 a 1] a 解説 a - a1 a1 a a, = ^ h a - a1 z- a1 z a1 a, = ^ h a - a1 0 z ] a1 ag z - a1, = 0 a - a1 7 素数平面1:重要例題

13 PMYA1-Z1J1-1 ポイント整理 - a = + i 1 a O 6 a 1+ i 6 着眼 1 a i ^cos i+i si ih a a b i ^cos i+ i si ih^a- bh + b 解答 w = cos + i si = + i 1 z 1 z1= wa 1 = c + im^ + ih 1 = + i+ i- = 1+ i 答 b = 1+ i z z = w^a- bh + b 1 = c + im" ^ + ih - ^1+ ih, + ^1+ ih 1 = c + im^ - 1h + ^1+ ih 1 = c im + ^1+ ih 5 1 = i 答 8

14 PMYA1-Z1J1-14 ポイント整理章複6 1 ^1+ ih 9 z z = - + i 着眼 1 a+ bi a b 1+ i 解答 i = c + im = ccos + i si m ^ ih = ccos + i si m = 64' cos c6$ m + i si c6$ m1= 64 答 - + i = c im - + i = ccos + i si m 4 4 z = ^cos i+ i si ih ^ 0 0 E i 1 h z i ^cos i + i si ih = ccos + i si m 4 4 = i = +k k 4 = ^ h 0 = i = + k 0 E i i = z 1 z z z1= ccos + i si m 4 4 z = ccos + i si m 1 1 z = ccos + i si m 1 1 右図答素数平面1:重要例題

2015－2017年度２次数学セレクション（複素数）解答解説

2015－2017年度２次数学セレクション（複素数）解答解説 05 次数学セレクション解答解説 [ 筑波大 ] ( + より, 0 となり, + から, ( (,, よって, の描く図形 C は, 点を中心とし半径がの円であるすなわち, 原点を通る円となる ( は虚数, は正の実数より, であるさて, w ( ( とおくと, ( ( ( w ( ( ( ここで, w は純虚数より, は純虚数となるすると, の描く図形 L は, 点を通り, 点と点