1. Prof : BEN ELKHATIR L’ensemble des nombres complexes ﺍﻟﺜﺎﻨﻴﺔ ﺒﺎﻜﺎﻟﻭﺭﻴﺎ ﻋﻠﻭﻡ ﺘﺠﺭﻴﺒﻴﺔ 30 ﻣﺠﻤﻮﻋﺔ اﻷﻋﺪاد اﻟﻌﻘﺪﻳﺔ
ﺨﺎﺼﻴﺔ10: • : -Iﺗﻘﺪﻳﻢ اﻟﻤﺠﻤﻮﻋﺔ
(a + ib )( a − ib ) = a 2
ﻟﺩﻴﻨﺎ : + b
2 2
ﻟﻜل ) ( a , bﻤﻥ 1(- ﻤﺒﺭﻫﻨﺔ ﻭ ﺘﻌﺭﻴﻑ:
ﺘﻭﺠﺩ ﻤﺠﻤﻭﻋﺔ ﻴﺭﻤﺯ ﻟﻬﺎ ﺒﺎﻟﺭﻤﺯ ﺘﺘﻀﻤﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ ﻭ ﺘﺤﺘﻭﻱ ﻋﻠﻰ ﻋﻨﺼﺭ ﻏﻴﺭ
ﻭ ﻟﻜل ﻋﺩﺩﻴﻥ ﻋﻘﺩﻴﻴﻥ z = a + ibﻭ ' z = a + ibﺒﺤﻴﺙ 0 ≠ zﻟﺩﻴﻨﺎ :
' ' '
ﺤﻘﻴﻘﻲ iﺒﺤﻴﺙ 1− = 2 iﻭ ﻟﻜل ﻋﻨﺼﺭ zﻤﻥ ﻴﻭﺠﺩ ﺯﻭﺝ ﻭﺤﻴﺩ ) ( a , bﻤﻥ 2 ﺒﺤﻴﺙ
' z aa ' + bb ' a 'b − ab
2' = ' . 2' + i . z = a + ib
z 2' a + b 2' a + b
• ﺇﺼﻁﻼﺤﺎﺕ:
ﺘﻤﺭﻴﻥ10: •
ﻋﻨﺎﺼﺭ ﺍﻟﻤﺠﻤﻭﻋﺔ ﺘﺴﻤﻰ ﺃﻋﺩﺍﺩﺍ ﻋﻘﺩﻴﺔ ﻭ ﺍﻟﺸﻜل z = a + ibﻴﺴﻤﻰ ﺍﻟﻜﺘﺎﺒﺔ ﺍﻟﺠﺒﺭﻴﺔ ﻟﻠﻌﺩﺩ
أ- ﺃﻜﺘﺏ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻌﻘﺩﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ :
ﺍﻟﻌﻘﺩﻱ ، zﺍﻟﻌﺩﺩ aﻴﺴﻤﻰ ﺍﻟﺠﺯﺀ ﺍﻟﺤﻘﻴﻘﻲ ل zﻭ bﻴﺴﻤﻰ ﺍﻟﺠﺯﺀ ﺍﻟﺘﺨﻴﻠﻲ ﻨﻜﺘﺏ :
) z 0 = ( 2 + 3i )( 3 − 4iﻭ ) z 1 = i (1 − 2iﻭ ) z 2 = ( 2 + i ) + (1 − 2i
3 3 3
) a = Re ( zﻭ ) . b = Im ( z
) (1 + i ) (1 − i
3 4 3
⎞ ⎛ 2 − 3i 5 + 3 3i ﺇﺫﺍ ﻜﺎﻥ 0 = aﻭ 0 ≠ bﻓﺈﻥ z = ibﻴﺴﻤﻰ ﻋﺩﺩﺍ ﻋﻘﺩﻴﺎ ﺘﺨﻴﻠﻴﺎ ﺼﺭﻓﺎ ، ﻭ ﻫﻭ ﻋﻨﺼﺭ ﻤﻥ
5. z = + ⎜= 4 zﻭ = 3 zﻭ ⎟ ﻭ
1− i ) (1 + i ⎠ ⎝ 3 + 2i ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺘﺨﻴﻠﻴﺔ ﺍﻟﺼﺭﻓﺔ * ، iﺇﺫﻥ : } * ∈ . i * = {ib / b
2
1 − 2 3i
ب- ﻟﻴﻜﻥ nﻋﺩﺩﺍ ﺼﺤﻴﺤﺎ ﻁﺒﻴﻌﻴﺎ ، ﺃﻜﺘﺏ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﺠﺒﺭﻱ i 4 nﻭ 1+ i 4 nﻭ 2 + i 4 n • ﺘﺴﺎﻭﻱ ﻋﺩﺩﻴﻥ ﻋﻘﺩﻴﻴﻥ:
و 3+ i 4 nﺛﻢ إﺳﺘﻨﺘﺞ 6002 iو 7002 iو 8002 iو 9002 . i z = z ⇔ Re ( z ) = Re ( z
' '
و) Im ( z ) = Im ( z '
) ﻟﻜل ) ' ( z , zﻤﻥ 2 ﻟﺩﻴﻨﺎ :
0102 9002
S 1 = ∑ iﻭ ) . S 2 = ∑ ( −i
k k
ج- ﺤﺩﺩ ﺍﻟﻤﺠﻤﻭﻋﻴﻥ : 1 Sﻭ 2 ، Sﺤﻴﺙ 0 = ) z = 0 ⇔ Re ( z ) = Im ( z ﻭ ﺒﺼﻔﺔ ﺨﺎﺼﺔ :
0= k 0= k
2(- ﺍﻟﺠﻤﻊ ﻭ ﺍﻟﻀﺭﺏ ﻓﻲ
:
3(- ﻤﺭﺍﻓﻕ ﻋﺩﺩ ﻋﻘﺩﻱ:
• ﺘﻌﺭﻴﻑ:
• ﺘﻌﺭﻴﻑ:
ﺒﺎﻟﻜﻴﻔﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ : ﺇﻟﻰ ﺍﻟﻤﺠﻤﻭﻋﺔ ﻴﻤﻜﻥ ﺘﻤﺩﻴﺩ ﻋﻤﻠﻴﺘﻲ ﺍﻟﺠﻤﻊ ﻭ ﺍﻟﻀﺭﺏ ﻓﻲ
، ﺍﻟﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ a − ibﻴﺴﻤﻰ ﻤﺭﺍﻓﻕ zﻭ ﻴﺭﻤﺯ ﻟﻪ ﻟﻴﻜﻥ z = a + ibﻋﻨﺼﺭﺍ ﻤﻥ
ﻨﻀﻊ : ﻟﻜل z = a + ibﻭ ' z = a + ibﻤﻥ
' '
ﺒﺎﻟﺭﻤﺯ ، zﺇﺫﻥ . z = a − ib
) ' z + z ' = ( a + a ' ) + i (b + bﻭ ) . z .z ' = ( aa ' − bb ' ) + i ( ab ' + a 'b
• ﻤﻠﺤﻭﻅﺔ:
• ﻤﻠﺤﻭﻅﺔ:
∈ ϕ : zﺘﻘﺎﺒل ∈z ﻟﺩﻴﻨﺎ : ، z = zﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﺍﻟﺘﻁﺒﻴﻕ ﻟﻜل zﻤﻥ
ﺘﺠﻤﻴﻌﻲ ، ﺘﺒﺎﺩﻟﻲ ﻭ ﻴﻘﺒل ﻋﻨﺼﺭﺍ ﻤﺤﺎﻴﺩﺍ ﻫﻭ 0 ﻭ ﻟﻜل ﻋﺩﺩ ﻋﻘﺩﻱ z = a + ib ﺍﻟﺠﻤﻊ ﻓﻲ
ﺘﻘﺎﺒﻠﻪ ﺍﻟﻌﻜﺴﻲ ﻨﻔﺴﻪ ، ﺒﻤﻌﻨﻰ ﺃﻥ : . ϕ = ϕ
1−
ﻤﻘﺎﺒل ﻫﻭ : . − z = −a − ib
ﺘﻤﺭﻴﻥ20: • ﻭ ﺍﻟﻀﺭﺏ ﻓﻲ }0{ − = * ﺘﺠﻤﻴﻌﻲ ، ﺘﺒﺎﺩﻟﻲ ﻭ ﻴﻘﺒل ﻋﻨﺼﺭﺍ ﻤﺤﺎﻴﺩﺍ ﻫﻭ 1 ﻭ ﻟﻜل ﻋﺩﺩ
. (E ) : z 2
ﺍﻟﻤﻌﺎﺩﻟﺔ : 0 = 4 + − 4z ﺤل ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ 1 a b
. 2 = 2 −i ﻋﻘﺩﻱ ﻏﻴﺭ ﻤﻨﻌﺩﻡ z = a + ibﻤﻘﻠﻭﺏ ﻫﻭ :
ﺨﺎﺼﻴﺔ20: • z a +b 2
2 a +b
z −z z +z ﺃﻱ ﺃﻥ : ﻭ ﺍﻟﻀﺭﺏ ﺘﻭﺯﻴﻌﻲ ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺠﻤﻊ ﻓﻲ
= ) . b = Im ( z = ) Re ( zﻭ ﻤﻬﻤﺎ ﻴﻜﻥ zﻤﻥ ﻟﺩﻴﻨﺎ :
2i 2 . 3
" z . ( z ' + z " ) = z .z ' + z .zﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻤﺜﻠﻭﺙ ) ( z , z , zﻤﻥ
' "
. z ∈i ﺇﺫﻥ : z ∈ ⇔ Im ( z ) = 0 ⇔ z = zﻭ ⇔ Re ( z ) = 0 ⇔ z = − z ﺠﺴﻡ ﺘﺒﺎﺩﻟﻲ . ( ﺒﻘﻭﻟﻨﺎ ﺃﻥ ).,+ , ﻨﻌﺒﺭ ﻋﻥ ﺨﺎﺼﻴﺎﺕ ﺍﻟﺠﻤﻊ ﻭ ﺍﻟﻀﺭﺏ ﻓﻲ
-1-
2. Prof : BEN ELKHATIR L’ensemble des nombres complexes ﺍﻟﺜﺎﻨﻴﺔ ﺒﺎﻜﺎﻟﻭﺭﻴﺎ ﻋﻠﻭﻡ ﺘﺠﺭﻴﺒﻴﺔ 30 ﻣﺠﻤﻮﻋﺔ اﻷﻋﺪاد اﻟﻌﻘﺪﻳﺔ
• ﺨﺎﺼﻴﺔ30: ﻟﻜل ﻋﺩﺩﻴﻥ ﻋﻘﺩﻴﻴﻥ 1 zﻭ 2 zﻟﺩﻴﻨﺎ :
ﻤﻬﻤﺎ ﺘﻜﻥ ﺍﻟﻤﺘﺠﻬﺘﺎﻥ 1 uﻭ 2 uﻭ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﺤﻘﻴﻘﻴﺎﻥ αﻭ βﻟﺩﻴﻨﺎ : ∈ αﻓﺈﻥ : 1 . α .z 1 = α .z 2 z 1 + z 2 = z 1 + zﻭ 2 z 1.z 2 = z 1.zﻭ ﺇﺫﺍ ﻜﺎﻥ
( ) ) ( ( ) ( ) ( )
2 aff u1 + u 2 = aff u1 + aff uﻭ 1aff α .u1 = α .aff u 1 ⎞ 1 ⎛
=⎟ ⎜ .
⎛z ⎞ z
ﻭ ﺇﺫﺍ ﻜﺎﻥ 0 ≠ 2 zﻓﺈﻥ : 1 = ⎟ 1 ⎜ ﻭ
(
1. aff α .u ﻭ ﺒﺼﻔﺔ ﻋﺎﻤﺔ : ) + β .u ) = α .aff (u ) + β .aff (u
2 1 2
2⎝z2 ⎠ z 2⎝z2 ⎠ z
) (
n
ﻭ ﻟﻜل ﻨﻘﻁﺘﻴﻥ Aﻭ Bﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻌﻘﺩﻱ ) ( Ρﻟﺩﻴﻨﺎ : . zn = z * *
ﻟﺩﻴﻨﺎ : ﻭ ﻟﻜل nﻤﻥ ﻭ ﺒﺼﻔﺔ ﺨﺎﺼﺔ ، ﻟﻜل zﻤﻥ
) ( ) ( ) (
) . aff AB = aff OB − aff OA = aff ( B ) − aff ( A -IIاﻟﺘﻤﺜﻴﻞ اﻟﻬﻨﺪﺳﻲ و ﻣﻌﻴﻠﺮ ﻋﺪد ﻋﻘﺪي:
• ﺨﺎﺼﻴﺔ40:
) 2 (e1 , eو اﻟﻤﺴﺘﻮى اﻟﻤﻮﺟﻪ ) ( Ρ اﻟﻤﺴﺘﻮى اﻟﻤﺘﺠﻬﻲ 2 Vﻣﻨﺴﻮب إﻟﻰ أﺳﺎس ﻣﺘﻌﺎﻣﺪ و ﻣﻤﻨﻈﻢ
ﺣﻴﺚ 0 ≠ α + β }) {( A , α ) ; ( B , β ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﻨﻘﻁﺔ Gﻤﺭﺠﺤﺎ ﻟﻠﻨﻅﻤﺔ ﺍﻟﻤﺘﺯﻨﺔ ( )
ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ و ﻣﻤﻨﻈﻢ 2 . O , e1 , e
) α .aff ( A ) + β .aff ( B
= ) . aff (G ﻓﺈن : ( )
) M ( x , yﺗﺸﻴﺮ إﻟﻰ اﻟﻨﻘﻄﺔ Mو زوج إﺣﺪاﺛﻴﺘﻴﻬﺎ ) ( x , yﻓﻲ اﻟﻤﻌﻠﻢ 2 . O , e1 , e
α +β
1(- ﺼﻭﺭﺓ ﻋﺩﺩ ﻋﻘﺩﻱ ﻭ ﻟﺤﻕ ﻨﻘﻁﺔ ﺃﻭ ﻤﺘﺠﻬﺔ:
) aff ( A ) + aff ( B
= ) . aff ( I ﺑﺼﻔﺔ ﺧﺎﺻﺔ إذا آﺎن Iهﻮ ﻣﻨﺘﺼﻒ ] [ ABﻓﺈن : • ﺘﻌﺭﻴﻑ10:
2 ﻟﻜل ﻋﺩﺩ ﻋﻘﺩﻱ z = x + iyﺍﻟﻨﻘﻁﺔ ) M ( x , yﺗﺴﻤﻰ ﺻﻮرة اﻟﻌﺪد z
• ﻤﻠﺤﻭﻅﺔ:
ﻭ ﻋﻜﺴﻴﺎ ، ﻟﻜل ﻨﻘﻁﺔ ) M ( x , yﻣﻦ ) ( Ρاﻟﻌﺪد اﻟﻌﻘﺪي z = x + iyﻴﺴﻤﻰ ﻟﺤﻕ ﺍﻟﻨﻘﻁﺔ M
) ( ) (
ﺍﻟﺭﺒﺎﻋﻲ ﺍﻟﺫﻱ ِﺅﻭﺴﻪ ﺍﻟﻨﻘﻁ ) M ( zﻭ M zﻭ ) M ( − zﻭ M − zﻤﺴﺘﻁﻴل ﻤﺭﻜﺯﻩ
ﺭ
z = aff ( Mأو ﺑﺎﺧﺘﺼﺎر ) . M ( z ) ﻧﻜﺘﺐ :
. −( ∪i ﻤﻥ ) ﺍﻟﻨﻘﻁﺔ Oﻤﻬﻤﺎ ﻴﻜﻥ z
ﻭ ﻴﻜﻭﻥ ﻫﺫﺍ ﺍﻟﻤﺴﺘﻁﻴل ﻤﺭﺒﻌﺎ ﺇﺫﺍ ﻭ ﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ : ) . Re ( z ) = Im ( z
و )(Ρ ( )
اﻟﻤﺴﺘﻘﻴﻢ 1 O , eﻳﺴﻤﻰ اﻟﻤﺤﻮر اﻟﺤﻘﻴﻘﻲ ، 2 O , eﻳﺴﻤﻰ اﻟﻤﺤﻮر اﻟﺘﺨﻴﻠﻲ ( )
ﻳﺴﻤﻰ اﻟﻤﺴﺘﻮى اﻟﻌﻘﺪي .
2(- ﻤﻌﻴﺎﺭ ﻋﺩﺩ ﻋﻘﺩﻱ:
ﻤﻠﺤﻭﻅﺔ: •
• ﺘﻌﺭﻴﻑ:
. z ∈i ∈ zﻭ ) ⇔ M ( z ) ∈ (Oy ﻟﺩﻴﻨﺎ : ) ⇔ M ( z ) ∈ (Ox
ﻟﻴﻜﻥ z = x + iyﻋﺩﺩﺍ ﻋﻘﺩﻴﺎ ﻭ Mﻨﻘﻁﺔ ﻤﻥ ) ( Ρﻟﺤﻘﻬﺎ z
∈z − ∈ zﻭ ) ' ⇔ M ( z ) ∈ ⎡Ox
⎣ + ﻭ ﺒﺼﻔﺔ ﺨﺎﺼﺔ : ) ⇔ M ( z ) ∈ [Ox
ﺍﻟﻌﺩﺩ ﺍﻟﻤﻭﺠﺏ OM = x + yﻴﺴﻤﻰ ﻤﻌﻴﺎﺭ ﺍﻟﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ zﻭ ﻴﺭﻤﺯ ﻟﻪ ﺒﺎﻟﺭﻤﺯ z 2 2
. z ∈i − z ∈ iﻭ ) ' ⇔ M ( z ) ∈ ⎡Oy
⎣ + ) ⇔ M ( z ) ∈ [Oy
. z = x +y 2 2
ﺇﺫﻥ :
• ﺘﻌﺭﻴﻑ20:
ﻤﻠﺤﻭﻅﺔ: • ﻟﻜل ﻋﺩﺩ ﻋﻘﺩﻱ z = x + iyﺍﻟﻤﺘﺠﻬﺔ ) u ( x , yﺘﺴﻤﻰ ﺼﻭﺭﺓ ﺍﻟﻌﺩﺩ ، zﻭ ﻋﻜﺴﻴﺎ ﻟﻜل ﻤﺘﺠﻬﺔ
2 ، z .z = x 2 + yﺇﺫﻥ : z = z .z ﻟﺩﻴﻨﺎ : ﻤﻬﻤﺎ ﻴﻜﻥ z = x + iyﻤﻥ
)(
) u ( x , yﻤﻥ 2 Vاﻟﻌﺪد اﻟﻌﻘﺪي z = x + iyﻴﺴﻤﻰ ﻟﺤﻕ uﻭ ﻨﻜﺘﺏ : . z = aff u
ﻭ ﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ ، ﻟﻜل ﻨﻘﻁﺘﻴﻥ Aﻭ Bﻤﻥ ) ( Ρﻟﺩﻴﻨﺎ : . AB = AB = z B − z A
. ﺇﺫﻥ ﻟﻜل ﻨﻘﻁﺔ Mﻣﻦ ) ( Ρﻟﺪﻳﻨﺎ : ) aff (OM ) = aff ( M
-2-
3. Prof : BEN ELKHATIR L’ensemble des nombres complexes ﺍﻟﺜﺎﻨﻴﺔ ﺒﺎﻜﺎﻟﻭﺭﻴﺎ ﻋﻠﻭﻡ ﺘﺠﺭﻴﺒﻴﺔ 30 ﻣﺠﻤﻮﻋﺔ اﻷﻋﺪاد اﻟﻌﻘﺪﻳﺔ
-IIIﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ﻟﻌﺩﺩ ﻋﻘﺩﻱ ﻏﻴﺭ ﻤﻨﻌﺩﻡ: ﺨﺎﺼﻴﺔ50: •
( )
ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻌﻘﺩﻱ ) ( Ρﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ و ﻣﺒﺎﺷﺮ 2 . O , e1 , e ﻟﺩﻴﻨﺎ : ﻤﻬﻤﺎ ﻴﻜﻥ zﻤﻥ
1(- ﻋﻤﺩﺓ ﻋﺩﺩ ﻋﻘﺩﻱ ﻏﻴﺭ ﻤﻨﻌﺩﻡ: z = − z = − z = zﻭ Re ( z ) ≤ Re ( z ) ≤ zﻭ . Im ( z ) ≤ Im ( z ) ≤ z
• ﺘﻌﺭﻴﻑ: . z = 1 ⇔ z −1 = z ﻭ 0= z =0⇔z
ﻟﻴﻜﻥ zﻋﺩﺩﺍ ﻋﻘﺩﻴﺎ ﻏﻴﺭ ﻤﻨﻌﺩﻡ ﻭ Mﻨﻘﻁﺔ ﻤﻥ ) ( Ρﻟﺤﻘﻬﺎ ، zﻜل ﻗﻴﺎﺱ ﻭ ﻟﻜل ﻋﺩﺩﻴﻥ ﻋﻘﺩﻴﻴﻥ 1 zﻭ 2 zﻟﺩﻴﻨﺎ :
e1 ,OMﻴﺴﻤﻰ ﻋﻤﺩﺓ ﺍﻟﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ zﻭ ﻴﺭﻤﺯ ﻟﻪ ﺒﺎﻟﺭﻤﺯ ) arg ( z ( ﻟﻠﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ ) . 2 z1 − z 2 ≤ z1 + z ﻭ 2 z1 + z 2 ≤ z1 + z ﻭ 2 z 1.z 2 = z 1 . z
(
. arg ( z ) ≡ e1 ,OM ﺇﺫﻥ : ] )[2π .
1z
2z
z
1 =
2z
ﻭ
1
2z
=
1
2z
ﻭ ﺇﺫﺍ ﻜﺎﻥ 0 ≠ 2 zﻓﺈﻥ :
• ﺤﺎﻻﺕ ﺨﺎﺼﺔ: ﺘﻤﺭﻴﻥ20: •
ﻤﻬﻤﺎ ﻴﻜﻥ zﻤﻥ * ﻟﺩﻴﻨﺎ :
) 3i ( 3 − 4i
2
∈z *
∈ zﻭ ] ⇔ arg ( z ) ≡ π [ 2π *
] ⇔ arg ( z ) ≡ 0 [ 2π = . z أ- ﺤﺩﺩ ﻤﻌﻴﺎﺭ ﺍﻟﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ zﺤﻴﺙ :
( () )
− + 3
5 − 2i 1 + 3i
∈. z *
ﻭ ﺒﺼﻔﺔ ﻋﺎﻤﺔ : ] ⇔ arg ( z ) ≡ 0 [π
3− z
π π ='. z ب- ﻟﻜل zﻤﻥ } − {2iﻨﻀﻊ :
z ∈i *
− − ≡ ) ⇔ arg ( z z ∈ iﻭ ] [ 2π *
+ ≡ ) ⇔ arg ( z ] [ 2π z − 2i
2 2
. z ∈i *
≡ ) ⇔ arg ( z
π
ﻭ ﺒﺼﻔﺔ ﻋﺎﻤﺔ : ] [π
) (O ,e ,e
1 2 ﺤﺩﺩ ﺜﻡ ﺃﻨﺸﻰﺀ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻌﻘﺩﻱ ) ( Ρﺍﻟﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻭ ﻤﻤﻨﻅﻡ
2 اﻟﻤﺠﻤﻮﻋﺎت اﻟﺘﺎﻟﻴﺔ :
• ﺨﺎﺼﻴﺔ60: ∈ ' ( Γ1 ) = {M ( z ) ∈ ( Ρ ) / zو } ( Γ 2 ) = {M ( z ) ∈ ( Ρ ) / z ∈ i
'
}
ﻜل ﻋﺩﺩ ﻋﻘﺩﻱ ﻏﻴﺭ ﻤﻨﻌﺩﻡ zﻴﻜﺘﺏ ﺒﻜﻴﻔﻴﺔ ﻭﺤﻴﺩﺓ ﻋﻠﻰ ﺸﻜل ) z = r ( cos θ + i sin θ
. ' ( Γ 3 ) = {M ( z ) ∈ ( Ρ ) / z 1= } و
ﺤﻴﺙ : r = zﻭ ] . θ ≡ arg ( z ) [ 2π
2z − i
ﻨﻜﺘﺏ ﺒﺎﺨﺘﺼﺎﺭ ] z = [ r , θﻭ ﺘﺴﻤﻰ ﻫﺫﻩ ﺍﻟﻜﺘﺎﺒﺔ ﺒﺎﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ﻟﻠﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ . z = ". z ج- ﻟﻜل zﻤﻥ − ﻨﻀﻊ :
z −z
ﺇﺫﺍ ﻜﺎﻥ z = x + iyﻓﺈﻥ :
ﻤﻠﺤﻭﻅﺔ: •
) (O ,e ,e
1 2 ﺤﺩﺩ ﺜﻡ ﺃﻨﺸﻰﺀ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻌﻘﺩﻱ ) ( Ρﺍﻟﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻭ ﻤﻤﻨﻅﻡ
اﻟﻤﺠﻤﻮﻋﺘﻴﻦ :
y x
= . sin θ
r
= cos θﻭ
r
2 r = x 2 + yﻭ θﻋﺩﺩ ﺤﻘﻴﻘﻲ ﺒﺤﻴﺙ : { "
}
( Σ1 ) = {M ( z ) ∈ ( Ρ ) / z " ∈ iﻭ 1 = . ( Σ1 ) = M ( z ) ∈ ( Ρ ) / z }
ﺨﺎﺼﻴﺔ70: • ﺘﻤﺭﻴﻥ30: •
⎤⎡ π ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ : ﺤل ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ
ﻟﻜل ) ( x , yﻤﻥ [∞+ ,0] × [∞+ ,0] ﻟﺩﻴﻨﺎ : ]0 , x = [ xﻭ ⎥ , . iy = ⎢ y
⎦2 ⎣ . (E2 ) : z (
ﻭ + 3.z = 2 + 3i z ) ( E1 ) : z 2 + 2 z
2
0 = 3−
-3-
4. Prof : BEN ELKHATIR L’ensemble des nombres complexes ﺍﻟﺜﺎﻨﻴﺔ ﺒﺎﻜﺎﻟﻭﺭﻴﺎ ﻋﻠﻭﻡ ﺘﺠﺭﻴﺒﻴﺔ 30 ﻣﺠﻤﻮﻋﺔ اﻷﻋﺪاد اﻟﻌﻘﺪﻳﺔ
ﺃﺤﺴﺏ 2 zﻭ ﺤﺩﺩ ﻤﻌﻴﺎﺭﻩ ﻭ ﻋﻤﺩﺘﻪ ، ﺜﻡ ﺇﺴﺘﻨﺘﺞ ﻤﻌﻴﺎﺭ ﻭ ﻋﻤﺩﺓ . z ⎡ ⎤π
ﻭ ﻟﻜل ) ( x , yﻤﻥ [0 ,∞−] × [0 ,∞−] ﻟﺩﻴﻨﺎ : ] x = [ − x , πﻭ ⎥ − , . iy = ⎢ − y
• ﺘﻤﺭﻴﻥ50: ⎣ ⎦2
أ- ﺃﻜﺘﺏ ﺘﺒﻌﺎ ﻟﻘﻴﻡ ﺍﻟﻌﺩﺩ θﻤﻥ ﺍﻟﻤﺠﺎل [ [ −π , πﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ﻟﻠﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ z θﺤﻴﺙ : • ﺨﺎﺼﻴﺔ80:
. z θ = 1 − cos θ + i sin θ ﻟﻜل ﻋﺩﺩﻴﻥ ﻋﻘﺩﻴﻴﻥ ﻏﻴﺭ ﻤﻨﻌﺩﻤﻴﻥ 1 zﻭ 2 zﻟﺩﻴﻨﺎ :
1 2 ⎧ z1 = z
⎪
= zα ب- ﺃﻜﺘﺏ ﺘﺒﻌﺎ ﻟﻘﻴﻡ ﺍﻟﻌﺩﺩ αﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ﻟﻠﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ z αﺤﻴﺙ : ⎨ ⇔ 2 z1 = z
1 + i tan α ] ⎪arg ( z 1 ) ≡ arg ( z 2 ) [ 2π
⎩
⎫⎧ π π
ﻭ ⎬ , −⎨ − [ . α ∈ [ −π , π ⎞ ⎛z
⎭2 2 ⎩ ] arg ( z 1.z 2 ) ≡ arg ( z 1 ) + arg ( z 2 ) [ 2πﻭ ] arg ⎜ 1 ⎟ ≡ arg ( z 1 ) − arg ( z 2 ) [ 2π
ج- ﺤﺩﺩ ﻤﺠﻤﻭﻋﺔ ﻗﻴﻡ ﺍﻟﻌﺩﺩ ﺍﻟﺼﺤﻴﺢ ﺍﻟﻨﺴﺒﻲ nﺍﻟﺘﻲ ﻴﻜﻭﻥ ﻤﻥ ﺃﺠﻠﻬﺎ ﻋﻠﻰ ﺍﻟﺘﻭﺍﻟﻲ : ⎠ 2⎝z
ﻭ ] arg ( z n ) ≡ n .arg ( z ) [ 2πﻟﻜل zﻤﻥ * ﻭ nﻤﻥ
( : )3 ( ) ( : )2 ( ) ( : )1( )
n n n .
. 3+i ∈i *
ﻭ 3 +i ∈ *
− ، 3 +i ∈ *
+
⎞1⎛
د- ﺤﺩﺩ ﺜﻡ ﺃﻨﺸﻰﺀ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻌﻘﺩﻱ ) ( Ρﺍﻟﻤﺠﻤﻭﻋﺔ ) ( Γﻟﻠﻨﻘﻁ ) M ( zﺍﻟﺘﻲ ﺘﺤﻘﻕ : ﻭﺒﺼﻔﺔ ﺨﺎﺼﺔ : ] arg ⎜ ⎟ ≡ − arg ( z ) [ 2πﻭ ] arg ( − z ) ≡ π + arg ( z ) [ 2π
⎠ ⎝z
π
− ≡ )1 + . arg ( z − 2i ] [ 2π
2
1 z
) (
ﻭ ﻟﺩﻴﻨﺎ ﺃﻴﻀﺎ : ] ) arg z ≡ − arg ( z ) [ 2πﻷن 2 = ( .
z z
2(- ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ ﻟﻤﺘﺠﻬﺘﻴﻥ ﻭ ﻋﻤﺩﺓ ﺨﺎﺭﺝ ﻟﺤﻘﻴﻬﻤﺎ:
• ﺨﺎﺼﻴﺔ90: • ﻤﻠﺤﻭﻅﺔ:
(e , AB ) ≡ arg ( z
1 B ﻟﻜل ﻨﻘﻁﺘﻴﻥ ﻤﺨﺘﻠﻔﺘﻴﻥ Aﻭ Bﻤﻥ ) ( Ρﻟﺩﻴﻨﺎ : ] − z A ) [ 2π
⎞ ⎛α
ﺇﺫﺍ ﻜﺎﻥ 0 > αﻓﺈﻥ :
] arg (α .z ) ≡ arg ( z ) [ 2πﻭ ] arg ⎜ ⎟ ≡ − arg ( z ) [ 2π
ﻭ ﻟﻜل ﺃﺭﺒﻊ ﻨﻘﻁ Aﻭ Bﻭ Cﻭ Dﻤﻥ ) ( Ρﺒﺤﻴﺙ : A ≠ Bﻭ C ≠ Dﻟﺩﻴﻨﺎ : ⎠ ⎝z
ﺃﻤﺎ ﺇﺫﺍ ﻜﺎﻥ 0 < αﻓﺈﻥ :
( AB ,CD ) ≡ arg ⎜⎝⎛ zz
⎞ − zC
⎠ B −zA
] ⎟ [ 2π
D
⎞ ⎛α
] arg (α .z ) ≡ π + arg ( z ) [ 2πﻭ ] . arg ⎜ ⎟ ≡ π − arg ( z ) [ 2π
ﻭ ﺒﺼﻔﺔ ﺨﺎﺼﺔ ﺇﺫﺍ ﻜﺎﻨﺕ Aﻭ Bﻭ Cﺜﻼﺙ ﻨﻘﻁ ﻤﺨﺘﻠﻔﺔ ﻤﺜﻨﻰ ﻤﺜﻨﻰ ﻤﻥ ) ( Ρﻓﺈﻥ : ⎠ ⎝z
• ﺘﻤﺭﻴﻥ40:
. ( AB , AC ) ≡ arg ⎛⎜⎝ zz C
B
⎞ −zA
] ⎟ [ 2π
⎠ −zA . sin
5π
21
cosو
5π
21
= 0 ، zﺛﻢ إﺳﺘﻨﺘﺞ
1+ i
3 −i
أ- ﺣﺪد ﻣﻌﻴﺎر و ﻋﻤﺪة اﻟﻌﺪد اﻟﻌﻘﺪي
ﺇﺴﺘﻨﺘﺎﺠﺎﺕ: • 3 +i 1 − 3i
ﻟﺘﻜﻥ Aﻭ Bﻭ Cﺜﻼﺙ ﻨﻘﻁ ﻤﻥ ) ( Ρﻤﺨﺘﻠﻔﺔ ﻤﺜﻨﻰ ﻤﺜﻨﻰ . = 2 ، zﺛﻢ = 1 zو ﺏ- أآﺘﺐ ﻋﻠﻰ اﻟﺸﻜﻞ اﻟﻤﺜﻠﺜﻲ اﻟﻌﺪدﻳﻦ اﻟﻌﻘﺪﻳﻴﻦ
4 4
zC − z A إﺳﺘﻨﺘﺞ ﻣﻌﻴﺎر و ﻋﻤﺪة اﻟﻌﺪدﻳﻦ اﻟﻌﻘﺪﻳﻴﻦ : 2 u = z 1 + zو 2 . v = z 1 − z
. ∈ *
ﺘﻜﻭﻥ ﺍﻟﻨﻘﻁ Aﻭ Bﻭ Cﻣﺴﺘﻘﻴﻤﻴﺔ إذا و ﻓﻘﻂ إذا آﺎن :
zB −zA
ج- أ- ﻨﻌﺘﺒﺭ ﺍﻟﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ 3 + 2 . z = 2 − 3 − i
-4-
5. Prof : BEN ELKHATIR L’ensemble des nombres complexes ﺍﻟﺜﺎﻨﻴﺔ ﺒﺎﻜﺎﻟﻭﺭﻴﺎ ﻋﻠﻭﻡ ﺘﺠﺭﻴﺒﻴﺔ 30 ﻣﺠﻤﻮﻋﺔ اﻷﻋﺪاد اﻟﻌﻘﺪﻳﺔ
: -IVاﻟﻤﻌﺎدﻻت ﻣﻦ اﻟﺪرﺟﺔ اﻟﺜﺎﻧﻴﺔ ﻓﻲ zC − z A
. ﻭﻴﻜﻭﻥ ﺍﻟﻤﺜﻠﺙ ABCﻗﺎﺌﻡ ﺍﻟﺯﺍﻭﻴﺔ ﻓﻲ Aﺇﺫﺍ ﻭ ﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ : * ∈ i
∈: a *
1(- ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ z 2 = aﺤﻴﺙ zB −zA
∈. a *
ﺤﻴﺙ ﻟﺘﻜﻥ Sﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ( E ) : z 2 = aﻓﻲ z −zA
. C ﻭﻴﻜﻭﻥ ﺍﻟﻤﺜﻠﺙ ABCﻗﺎﺌﻤﺎ ﻭ ﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻥ ﺭﺃﺴﻪ Aﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ: = ±i
{
S = − a, a - ﺇﺫﺍ ﻜﺎﻥ ∈ aﻓﺈﻥ : } *
+
zB −zA
⎤ z −zA ⎡ π
∈ aﻓﺈﻥ : 0 = ) = ( i −a ) ⇔ ( z + i −a )( z − i −a ﻭﻴﻜﻭﻥ ﺍﻟﻤﺜﻠﺙ ABCﻤﺘﺴﺎﻭﻱ ﺍﻷﻀﻼﻉ ﺇﺫﺍ ﻭ ﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ : ⎥ ± ,1 =
2
2 (E ) ⇔ z . C
⎢ zB −zA
*
- ﺇﺫﺍ ﻜﺎﻥ ⎦3
− ⎣
ﺇﺫﻥ : } . S = {−i −a , i −a • ﺘﻤﺭﻴﻥ60:
أ- ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ) A ( 2 + 3iﻭ ) B ( 4 − iﻭ ) . C (10 + 2i
ﻨﻘﻭل ﺇﻥ ﺍﻟﻌﺩﺩﻴﻥ ﺍﻟﻌﻘﺩﻴﻴﻥ z 1 = −i −aﻭ z 2 = i −aﻫﻤﺎ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﻤﺭﺒﻌﻴﻥ
ﻟﻠﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ﺍﻟﺴﺎﻟﺏ . a ﺒﻴﻥ ﺃﻥ ﺍﻟﻤﺜﻠﺙ ABCﻗﺎﺌﻡ ﺍﻟﺯﺍﻭﻴﺔ ﻓﻲ . B
- ﺇﺫﺍ ﻜﺎﻥ − ∈ aﺃﻱ ﺃﻥ ∉ ، a ب- ﺑﻴﻦ أن اﻟﻤﻌﺎدﻟﺔ 0 = 1 + ( E ) : z 2 − 2zﺗﻘﺒﻞ ﻓﻲ اﻟﻤﺠﻤﻮﻋﺔ ﺛﻼث ﺣﻠﻮل ، و ﺣﺪد
× ∈ ) (α , βﻭ z = x + iy *
ﻨﻀﻊ : a = α + i βﺤﻴﺙ .ﻃﺒﻴﻌﺔ اﻟﻤﺜﻠﺚ ABCﺍﻟﺫﻱ ﺃﻟﺤﺎﻕ ﺭﺅﻭﺴﻪ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ) ( E
) ( E ) ⇔ ( x + iy
2
ﺇﺫﻥ : = α + i β ⇔ x 2 − y 2 + 2ixy = α + i β ج- ﺤﺩﺩ ﺜﻡ ﺃﻨﺸﻰﺀ ﺍﻟﻤﺠﻤﻭﻋﺔ ) ( Σﻟﻠﻨﻘﻁ ) M ( zﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻌﻘﺩﻱ ) ( Ρﺍﻟﺘﻲ ﺘﻜﻭﻥ ﻤﻥ ﺃﺠﻠﻬﺎ
⎧x 2 − y 2 = α ﺍﻟﻨﻘﻁ ) M ( zﻭ ) N ( i zﻭ ) P ( −3iﻤﺴﺘﻘﻴﻤﻴﺔ .
. ⎨ ⇔ ) (E ﺃﻱ ﺃﻥ :
⎩2xy = β ﺨﺎﺼﻴﺔ01: •
zﺃﻱ ﺃﻥ : 2 . x 2 + y 2 = α 2 + β
2
ﻭ ﺒﻤﺎ ﺃﻥ z 2 = aﻓﺈﻥ : = a ﻟﻴﻜﻥ ABCDﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ ) هﺬا ﻳﻌﻨﻲ أن : . ( z B − z A = z C − z D
z −zA
2 ⎧(1) : x 2 + y 2 = α 2 + β . D ﻳﻜﻮن ABCDﻤﺴﺘﻁﻴﻼ ﺇﺫﺍ ﻭ ﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ : * ∈ i
⎪ zB −zA
⎪
. ( E ) ⇔ ⎨( 2 ) : x 2 − y 2 = α ﻭ ﺒﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ : z −zB
⎪ . D ﻭ ﻴﻜﻭﻥ ABCDﻤﻌﻴﻨﺎ ﺇﺫﺍ ﻭ ﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ : * ∈ i
⎪( 3) : 2xy = β
⎩
zC − z A
z −zB z −zA
∈ βﻭ ﺃﻥ ﺇﺸﺎﺭﺘﻪ ﺘﺤﺩﺩ ﻤﺎ ﺇﺫﺍ ﻜﺎﻥ ل xﻭ yﻨﻔﺱ ﺍﻹﺸﺎﺭﺓ ﺃﻭ ﺇﺸﺎﺭﺓ ﻤﺨﺘﻠﻔﺔ . *
ﻻﺤﻅ ﺃﻥ . D Dﻭ * ∈i ﻭ ﻴﻜﻭﻥ ﻤﺭﺒﻌﺎ ﺇﺫﺍ ﻭ ﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ : * ∈ i
zC − z A zB −zA
ﻨﺴﺘﻌﻤل ﺍﻟﻤﻌﺎﺩﻟﺘﻴﻥ )1( ﻭ ) 2 ( ﻓﻨﺤﺼل ﻋﻠﻰ :
ﺘﻤﺭﻴﻥ70: •
= 2y
2 −α + α 2 + β
2
= 2 xﻭ 0>
2α + α2 + β
2
0> . C ( )
Bﻭ 3 −i ( )
أ- ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ) A ( 2iﻭ 3 + i
ﺒﻴﻥ ﺃﻥ ﺍﻟﺭﺒﺎﻋﻲ OABCﻤﻌﻴﻥ .
2 −α + α 2 + β 2α + α2 + β ب- ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ) A ( −2 + 5iﻭ ) B ( 3 + 2iﻭ ) C ( 6 + 7iﻭ ) . D (1 + 10i
±= y ±= xﻭ ﺇﺫﻥ :
2 2 ﺒﻴﻥ ﺃﻥ ﺍﻟﺭﺒﺎﻋﻲ ABCDﻤﺭﺒﻊ .
ﺜﻡ ﻨﺴﺘﻌﻤل ﺍﻟﻤﻌﺎﺩﻟﺔ )3 ( : 0 < β < 0 ⇒ xyﻭ 0 > . β > 0 ⇒ xy
-5-
6. Prof : BEN ELKHATIR L’ensemble des nombres complexes ﺍﻟﺜﺎﻨﻴﺔ ﺒﺎﻜﺎﻟﻭﺭﻴﺎ ﻋﻠﻭﻡ ﺘﺠﺭﻴﺒﻴﺔ 30 ﻣﺠﻤﻮﻋﺔ اﻷﻋﺪاد اﻟﻌﻘﺪﻳﺔ
∈ ) : (b , c 2
∈ aﻭ *
2(- ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ 0 = ( E ) : az 2 + bz + cﺤﻴﺙ • ﻤﺒﺭﻫﻨﺔ:
⎛⎡ ⎞ b
2
⎤ ∆ ﻟﻜل ﻋﺩﺩ ﻋﻘﺩﻱ ﻏﻴﺭ ﻤﻨﻌﺩﻡ aﺠﺫﺭﻴﻥ ﻤﺭﺒﻌﻴﻥ ﻤﺘﻘﺎﺒﻼﻥ ﻫﻤﺎ ﺤﻠﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ( E ) : z = a
2
ﻤﻬﻤﺎ ﻴﻜﻥ zﻤﻥ ﻟﺩﻴﻨﺎ : ⎥ 2 − ⎟ + az + bz + c = a ⎢⎜ zﺤﻴﺙ ∆ = b − 4ac
2 2
. ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ
⎝⎢
⎣ ⎥ 2a ⎠ 4a ⎦ • ﺃﻤﺜﻠﺔ:
b
−= 0. z ﺇﺫﺍ ﻜﺎﻥ 0 = ∆ ﻓﺎﻟﻤﻌﺎﺩﻟﺔ ) ( Eﺘﻘﺒل ﺤﻼ ﻭﺤﻴﺩﺍ - ﻟﻠﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ﺍﻟﺴﺎﻟﺏ 4− = aﺠﺫﺭﻴﻥ ﻤﺭﺒﻌﻴﻥ ﻋﻘﺩﻴﻴﻥ ) ﺘﺨﻴﻠﻴﻴﻥ ﺼﺭﻓﻴﻥ ( هﻤﺎ :
2a z 1 = −2iو . z 2 = 2i
ﻭ ﺇﺫﺍ ﻜﺎﻥ 0 ≠ ∆ ﻓﺈﻨﻪ ﻴﻘﺒل ﺠﺫﺭﻴﻥ ﻤﺭﺒﻌﻴﻥ δﻭ ) −δﻋﻘﺪﻳﻴﻦ ﻓﻲ ﺣﺎﻟﺔ + − ∈ ∆ (
اﻟﺠﺬرﻳﻦ اﻟﻤﺮﺑﻌﻴﻦ ﻟﻠﻌﺪد 8− هﻤﺎ : z 1 = −2 2iو . z 2 = 2 2i
⎛⎡ ⎤ ⎞ b ⎞ ⎛δ
2 2
ﻭ ﻟﺩﻴﻨﺎ : ⎥ ⎟ ⎜ − ⎟ + . az + bz + c = a ⎢⎜ z
2
- ﻟﻨﺤﺩﺩ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﻤﺭﺒﻌﻴﻥ ﻟﻠﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ : a = 4 − 3i
⎝⎢
⎣ ⎥ ⎠ 2a ⎠ ⎝ 2a ⎦ 2 ) 3− ( + 2 4 = 2 ⎧ x 2 + y
⎛ ⎛ ⎞ −b − δ ⎞ −b + δ ⎪
− (E ) ⇔ a ⎜ z − ⎟⎜ z ﺇﺫﻥ : 0 = ⎟ ⎪
4 = 2 z = 4 − 3i ⇔ ⎨x 2 − y
2
ﻨﻀﻊ : ، z = x + iyﺇﺫﻥ :
⎝ ⎝ ⎠ 2a ⎠ 2a
−b + δ −b − δ 3− = ⎪2xy
= 2. z = 1 zﻭ ﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ ) ( Eﺘﻘﺒل ﺤﻠﻴﻥ ﻤﺨﺘﻠﻔﻴﻥ ﻫﻤﺎ : ⎪
⎩
2a 2a
• ﻤﻠﺨﺹ: ﺒﻤﻌﻨﻰ ﺃﻥ : 9 = 2 2xﻭ 1 = 2 2 yﻭ 0 < . xy
ﻟﺘﻜﻥ Sﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ : 0 = ( E ) : az + bz + cﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ
2 ⎞2 2 3 ⎛ 2 3⎛ ⎞2
. − ⎜ = ) (x , y
⎜ , ⎜ = ) ( x , yﺃﻭ ⎟
⎟ ﺃﻱ : ⎟ 2 − , 2 ⎜
⎟
ﻭ ∆ = b 2 − 4acﻤﻤﻴﺯﻫﺎ . ⎝ 2 ⎠ 2 ⎝ ⎠
⎫ ⎧ b 2 2
ﺇﺫﺍ ﻜﺎﻥ 0 = ∆ ﻓﺈﻥ : ⎬ −⎨ = . S
⎭ ⎩ 2a
= 2. z ﺇﺫﻥ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﻤﺭﺒﻌﻴﻥ ل a = 4 − 3iﻫﻤﺎ : ) z 1 = ( 3 − iﻭ ) ( −3 + i
2 2
⎫ ⎧ −b − δ −b + δ - ﻟﺘﺤﺩﻴﺩ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﻤﺭﺒﻌﻴﻥ ﻟﻠﻌﺩﺩ ، a = 1 + 3iﻨﻜﺘﺏ aﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ :
⎨ = ، Sﺤﻴﺙ δﻫﻭ ﺃﺤﺩ ﺠﺫﺭﻱ ﺍﻟﻤﻤﻴﺯ ∆ . , ﻭ ﺇﺫﺍ ﻜﺎﻥ 0 ≠ ∆ ﻓﺈﻥ ⎬
⎩ 2a ⎭ 2a ⎤⎡ π ⎤⎡ π
⎥ ,2 ⎢ = aﻭ ﻨﻀﻊ : ] ، z = [ r , θﺇﺫﻥ : ⎥ ,2 ⎢ = ⎤ z 2 = a ⇔ ⎡ r 2 , 2θ
⎣ ⎦
• ﻤﻠﺤﻭﻅﺔ: ⎦3 ⎣ ⎦3 ⎣
ﺇﺫﺍ ﻜﺎﻨﺕ ﻤﻌﺎﻤﻼﺕ ﺍﻟﻤﻌﺎﺩﻟﺔ ) ( Eﺃﻋﺩﺍﺩ ﺤﻘﻴﻘﻴﺔ ﻭ ﻜﺎﻥ 0 < ∆ ﻓﺈﻥ ﺃﺤﺩ ﺠﺫﺭﻱ ∆ π π
≡. θ ] [π ≡ 2θﺃﻱ ﺃﻥ : 2 = rﻭ ] [ 2π ﻭ ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ : 2 = 2 rﻭ
ﻫﻭ ∆− δ = iﻭ ﺤﻠﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ) ( Eﻋﺩﺩﻴﻥ ﻋﻘﺩﻴﻴﻥ ﻤﺘﺭﺍﻓﻘﻴﻥ ﻭ ﻫﻤﺎ : 6 3
⎡ ⎤ 5π ⎡ ⎤π
∆− −b + i ∆− −b − i ﺇﺫﻥ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﻤﺭﺒﻌﻴﻥ ل aﻫﻤﺎ : ⎥ ,2 ⎢ = 1 zﻭ ⎥ − ,2 ⎢ = 2 z
= 2 ( z 2 = z1 ) z = 1 zﻭ ⎣ ⎦ 6 ⎣ ⎦6
2a 2a
ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ : ﺤل ﻓﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻌﻘﺩﻴﺔ
ﺘﻤﺭﻴﻥ90: • −= 2. z
2
2
( )
= 1 zﻭ 3 +i
2
2
(
3+i )
0 = 982 + (1) : z 2 − 30zﻭ 0 = ( 2 ) : z − 6z + 9 − 6i
2 • ﺘﻤﺭﻴﻥ80: ﺤﺩﺩ ﺍﻟﺠﺫﺭﻴﻥ ﺍﻟﻤﺭﺒﻌﻴﻥ ﻟﻠﻌﺩﺩﻴﻥ ﺍﻟﻌﻘﺩﻴﻴﻥ : u = 1 − 2 3i
ﻭ . v = −5 + 12i
-6-
7. Prof : BEN ELKHATIR L’ensemble des nombres complexes ﺍﻟﺜﺎﻨﻴﺔ ﺒﺎﻜﺎﻟﻭﺭﻴﺎ ﻋﻠﻭﻡ ﺘﺠﺭﻴﺒﻴﺔ 30 ﻣﺠﻤﻮﻋﺔ اﻷﻋﺪاد اﻟﻌﻘﺪﻳﺔ
ﻟﻨﻨﺸﺭ ﻤﺜﻼ ) 2 ( z 1 + zﻭ ) 2 ( z 1 + zﺒﺎﺴﺘﻌﻤﺎل ﻤﺜﻠﺙ ﺒﺎﺴﻜﺎل . ⎛ ⎞ i
5 4
0 = ) ( 4 ) : (1 + i ) z 2 − 3z + 2 (1 − i ﻭ + 1⎜ − ( 3) : iz 2 + ( 2i − 1) z 0=⎟
⎝ ⎠4
ﻨﺤﺼل ﻋﻠﻰ ﻤﻌﺎﻤﻼﺕ ﻨﺸﺭ ) 2 ( z 1 + zﻭ ) 2 ( z 1 + zﺇﻨﻁﻼﻗﺎ ﻤﻥ ﺍﻟﺴﻁﺭ ﺍﻟﺨﺎﻤﺱ ﻭ ﺍﻟﺴﺎﺩﺱ
5 4
ﻋﻠﻰ ﺍﻟﺘﻭﺍﻟﻲ : . ( 5) : z 2
(
ﻭ 0 = − 2i .z + 1 + 2 3i)
ﺇﺫﻥ : 4 2 ( z 1 + z 2 ) = z 14 + 4z 13 .z 2 + 6z 12 .z 2 2 + 4z 1.z 23 + z
4
• ﺘﻤﺭﻴﻥ01:
أ- ﻟﺘﻜﻥ 1 Sﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ : 0 = 61 + ، ( E 1 ) : zﻭ 2 Sﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل
4
52 ( z 1 + z 2 ) = z 15 + 5z 14 .z 2 + 10z 13 .z 2 2 + 10z 12 .z 23 + 5z 1.z 2 4 + z
5
. ﻭ
ﺍﻟﻤﻌﺎﺩﻟﺔ : 0 = 61 + ) ( E 2 ) : ( z − 2i
4
ﺼﻴﻐﺔ ﺍﻟﺤﺩﺍﻨﻴﺔ: • .
ﺒﺤﻴﺙ 2 ≥ nﻭ ﻟﻜل ﻋﺩﺩﻴﻥ ﻋﻘﺩﻴﻴﻥ 1 zﻭ 2 zﻟﺩﻴﻨﺎ : ﻟﻜل nﻤﻥ ﺤﺩﺩ 1 Sﺜﻡ ﺇﺴﺘﻨﺘﺞ 2 ، Sﺜﻡ ﺒﻴﻥ ﺼﻭﺭ 2 Sﻓﻲ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻌﻘﺩﻱ ) ( Ρﺭﺅﻭﺱ ﻟﻤﺭﺒﻊ ﻴﺘﻡ ﺇﻨﺸﺎﺌﻪ .
n n
ﺍﻟﻤﻌﺎﺩﻟﺔ : 0 = 1 + ( E 3 ) : z 4 − 2z cos θﺤﻴﺙ . 0 < θ < π ب- ﺤل ﻓﻲ ﺍﻟﻤﺠﻤﻭﻋﺔ
( z 1 − z 2 ) = ∑ ( −1) C nk z 1n −k .z 2 k ( z 1 + z 2 ) = ∑C nk z 1n −k .z 2 k
n k n
ﻭ
0= k 0= k ج- ﺒﻴﻥ ﺃﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ : 0 = ( E ) : z − 2 ( λ cos θ + i sin θ ) z + 1 − λﺘﻘﺒل ﺤﻠﻴﻥ
2 2
Cﻫﻲ ﻤﻌﺎﻤﻼﺕ ﺍﻟﺴﻁﺭ 1 + nﻓﻲ ﻤﺜﻠﺙ ﺒﺎﺴﻜﺎل . k
n ﺤﻴﺙ ﺍﻷﻋﺩﺍﺩ ﻤﺨﺘﻠﻔﻴﻥ 1 zﻭ 2 zﻴﺘﻡ ﺘﺤﺩﻴﺩﻫﻤﺎ ) ﻟﻜل ) ( λ , θﻤﻥ × * ( ، ﺛﻢ أآﺘﺐ 1 zﻭ 2 z
ﺘﻤﺭﻴﻥ11: • ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ .
)1 − ( z )1 + ( z ﻭ ) ( z − 2i ﺃﻨﺸﺭ ) ( z + 2i
01 01 6 7
. ﻭ ﻭ -Vﺻﻴﻐﺔ ﻣﻮاﻓﺮ - اﻟﺘﺮﻣﻴﺰ اﻷﺳﻲ ﻟﻌﺪد ﻋﻘﺪي ﻏﻴﺮ ﻣﻨﻌﺪم:
2(- ﺼﻴﻐﺔ ﻤﻭﺍﻓﺭ: 1(- ﺼﻴﻐﺔ ﺍﻟﺤﺩﺍﻨﻴﺔ ﻭ ﻤﺜﻠﺙ ﺒﺎﺴﻜﺎل:
) 2 (z 1 + z
2
• ﺨﺎﺼﻴﺔ11: ﻤﻬﻤﺎ ﻴﻜﻥ ﺍﻟﻌﺩﺩﻴﻥ ﺍﻟﻌﻘﺩﻴﻴﻥ 1 zﻭ 2 zﻟﺩﻴﻨﺎ : 2 2 = z 12 + z 1.z 2 + z
) ( cos θ + i sin θ ﻟﺩﻴﻨﺎ : = cos nθ + i sin nθ ﻟﻜل θﻤﻥ ﻭ 32 . ( z 1 + z 2 ) = z 13 + 3z 12 .z 2 + 3z 1.z 2 2 + z
n 3
. ﻭ ﻟﻜل nﻤﻥ
ﺘﻤﺭﻴﻥ21: • ﻫﺎﺘﻴﻥ ﺍﻟﻤﺘﻁﺎﺒﻘﺘﻴﻥ ﻴﻤﻜﻥ ﺘﻌﻤﻴﻤﻬﻤﺎ ، ﺒﺎﺴﺘﻌﻤﺎل ﺠﺩﻭل ﻴﺴﻤﻰ ﻤﺜﻠﺙ ﺒﺎﺴﻜﺎل :
073− 7002
⎛ π ⎞π ⎛ π ⎞π
⎟ . ⎜ − cos + i sin ⎟ ⎜ cos + i sinﻭ ﺃﺤﺴﺏ : 1
⎝ 5 ⎠5 ⎝ 9 ⎠9
• ﻤﻠﺤﻭﻅﺔ: 1−1
ﺇﺫﺍ ﻜﺎﻥ : ) z = r ( cos θ + i sin θﻓﺈﻥ : ) ∀n ∈ : z = r ( cos nθ + i sin nθ
n n 1− 2 −1
ﻤﺜﺎل: • 1− 3 − 3 −1
9−
ﺃﻜﺘﺏ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﻤﺜﻠﺜﻲ ﺍﻟﻌﺩﺩ 3 ، z = 243 + iﺜﻡ ﺇﺴﺘﻨﺘﺞ . z
6 3 1− 4 − 6 − 4 −1
ﺤﺴﺎﺏ cos nθﻭ sin nθﺤﻴﺙ ∈ nﻭ 2 ≥ : n ﺘﻁﺒﻴﻘﺎﺕ: • 1 − 5 − 01 − 01 − 5 − 1
ﻟﺩﻴﻨﺎ ﻤﻥ ﺠﻬﺔ : ) ( cos θ + i sin θ ) = cos nθ + i sin nθﺻﻴﻐﺔ ﻣﻮاﻓﺮ ( 1 − 6 − 51 − 02 − 51 − 6 − 1
n
n
ﻭ ﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ : ) ( cos θ + i sin θ ) = ∑ i k C nk cos n − k θ sin k θﺻﻴﻐﺔ اﻟﺤﺪاﻧﻴﺔ (
n
ﻭ ﻻﺤﻅ ﺃﻨﻪ ﻓﻲ ﻨﺸﺭ ﺍﻟﻤﺘﻁﺎﺒﻘﺘﻴﻥ ﺍﻟﺴﺎﺒﻘﺘﻴﻥ ﺇﺴﺘﻌﻤﻠﺕ ﻤﻌﺎﻤﻼﺕ ﺍﻟﺴﻁﺭ ﺍﻟﺜﺎﻟﺙ ﻭ ﺍﻟﺭﺍﺒﻊ
0= k
ﻋﻠﻰ ﺍﻟﺘﻭﺍﻟﻲ .
-7-
8. Prof : BEN ELKHATIR L’ensemble des nombres complexes ﺍﻟﺜﺎﻨﻴﺔ ﺒﺎﻜﺎﻟﻭﺭﻴﺎ ﻋﻠﻭﻡ ﺘﺠﺭﻴﺒﻴﺔ 30 ﻣﺠﻤﻮﻋﺔ اﻷﻋﺪاد اﻟﻌﻘﺪﻳﺔ
ﺜﻡ ﺇﺴﺘﻨﺘﺞ ﺍﻟﻜﺘﺎﺒﺔ ﺍﻷﺴﻴﺔ ﻟﻸﻋﺩﺍﺩ ﺍﻟﻌﻘﺩﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ : ﻭ ﻤﻨﻬﻤﺎ ﻤﻌﺎ ﻨﺴﺘﻨﺘﺞ ﺃﻥ :
( )
n
2 1 ، sin nθ = Im ( Zﺤﻴﺙ . Z = ∑ i k C nk cos n − k θ sin k θ
+ 1 = 1 zﻭ ) . z 2 = 2 + 2 (1 + i z 0 = 3 + 3iﻭ )1 − ( i ) cos nθ = Re ( Zﻭ )
2 2 0= k
ﺇﺨﻁﺎﻁ cos n θﻭ sin θﺤﻴﺙ 2 ≥ : n
n
• ﻭ ﻻﺤﻅ ﺃﻨﻪ ﻴﺠﺏ ﺇﺴﺘﻌﻤﺎل ﻤﺜﻠﺙ ﺒﺎﺴﻜﺎل ﻟﺘﺤﺩﻴﺩ ﺍﻟﻤﻌﺎﻤﻼﺕ C nkﺤﻴﺙ . 0 ≤ k ≤ n
ﻴﻬﺩﻑ ﺍﻹﺨﻁﺎﻁ ﺇﻟﻰ ﺍﻟﺘﻌﺒﻴﺭ ﻋﻥ cos n θﻭ sin n θﺒﺩﻻﻟﺔ cos k θﻭ sin k θ • ﺘﻤﺭﻴﻥ31:
) ﺤﻴﺙ ( 1 ≤ k ≤ nو ذﻟﻚ ﺑﺎﺳﺘﻌﻤﺎل ﺻﻴﻐﺘﺎ أوﻟﻴﺮ ﺛﻢ ﺻﻴﻐﺔ ﺣﺪاﻧﻴﺔ ﻧﻴﻮﺗﻦ . ﺒﻴﻥ ﺃﻥ : cos 5θ = cos θ − 10 cos θ sin θ + 5cos θ sin θ
5 3 2 4
ﺃﻤﺜﻠﺔ: • ﻭ . sin 5θ = sin 5 θ − 10sin 3 θ cos 2 θ + 5sin θ cos 4 θ
ﺇﺨﻁﺎﻁ : sin θﻭ cos θﻭ sin θ
5 5 4
3(- ﺍﻟﺘﺭﻤﻴﺯ ﺍﻷﺴﻲ ﻟﻌﺩﺩ ﻋﻘﺩﻱ ﻏﻴﺭ ﻤﻨﻌﺩﻡ:
ﻨﻌﻠﻡ ﺃﻨﻪ ﻟﻜل ﻋﺩﺩ ﻋﻘﺩﻱ ﻏﻴﺭ ﻤﻨﻌﺩﻡ zﻴﻭﺠﺩ ﺯﻭﺝ ﻭﺤﻴﺩ ) ( r , θﻤﻥ × + ﺒﺤﻴﺙ :
4 *
⎞ ⎛ e i θ − e −i θ 1 4i θ
⎜ = sin θ ⎟ = (e − 4e + 6 − 4e ﻟﺪﻳﻨﺎ : ) + e −4i θ
4 2i θ −2 i θ
⎝ 2i 61 ⎠ ) ، z = r ( cos θ + i sin θﺤﻴﺙ r = zﻭ ] . θ ≡ arg ( z ) [ 2π
1 1 3 1 ﻨﻀﻊ : e i θ = cos θ + i sin θ
إذن : . sin 4 θ = ( 2 cos 4θ − 8cos 2θ + 6 ) = + cos 4θ − cos 2θ
61 8 8 2 ﻓﻴﺼﺒﺢ ﻟﺩﻴﻨﺎ : z = re i θﻭ ﺘﺴﻤﻰ ﻫﺫﻩ ﺍﻟﻜﺘﺎﺒﺔ ﻜﺘﺎﺒﺔ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻷﺴﻲ ﻟﻠﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ . z
5
⎞ ⎛ e i θ + e −i θ 1 5i θ • ﺃﻤﺜﻠﺔ:
⎜ = cos θ ⎟ = (e + 5e + 10e + 10e + 5e ﻭ ) + e −5i θ
5 3i θ iθ −i θ −3i θ
π π π
⎝ 2 23 ⎠ i −i −i
. 3 − 3i = 2 3.e 6
1 − i = 2.eﻭ 4
−i = eﻭ −1 = e i πﻭ2
1 5 5
. cos5 θ = cos 5θ + cos 3θ + cos θ ﺇﺫﻥ : ﺨﺎﺼﻴﺔ21: •
61 61 8 2i θ 1i θ
5
z 1 = r1.eﻭ z 2 = r2 .e ﻟﻜل ﻋﺩﺩﻴﻥ ﻋﻘﺩﻴﻴﻥ ﻏﻴﺭ ﻤﻨﻌﺩﻤﻴﻥ
⎞ ⎛ e i θ − e −i θ 1
⎜ = sin 5 θ
2i
= ⎟ ﻭ ) (e 5i θ − 5e 3i θ + 10e i θ − 10e −i θ + 5e −3i θ − e −5i θ ) 2z 1 r1 i (θ1 −θ
= .e ﻟﺩﻴﻨﺎ : ) 2 z 1.z 2 = r1r2 .e i (θ1 +θﻭ
⎝ ⎠ 32i 2z 2 r
1 5 5 ﻭ ﻟﻜل z = r .e i θﻤﻥ * ﻭ nﻤﻥ ﻟﺩﻴﻨﺎ : . z n = r n .e inθ
. sin 5 θ = sin 5θ − sin 3θ + sin θ ﺇﺫﻥ :
61 61 8 4(- ﺼﻴﻐﺘﺎ ﺃﻭﻟﻴﺭ ﻭ ﺇﺨﻁﺎﻁ cos n θﻭ sin n θﺤﻴﺙ 2 ≥ : n
• ﻤﻠﺤﻭﻅﺔ:
• ﺨﺎﺼﻴﺔ31: ) ﺻﻴﻐﺘﺎ أوﻟﻴﺮ (
ﺍﻟﻌﻤﻠﻴﺔ ﺍﻟﺘﻲ ﻨﺤﺴﺏ ﻓﻴﻬﺎ cos nθﻭ sin nθﺒﺎﺴﺘﻌﻤﺎل ﺼﻴﻐﺔ ﻤﻭﺍﻓﺭ ﻫﻲ ﻋﻤﻠﻴﺔ ﻋﻜﺴﻴﺔ iθ −i θ iθ −i θ
e −e e +e
ﻟﻠﻌﻤﻠﻴﺔ ﺍﻟﺘﻲ ﻴﺘﻡ ﻓﻴﻬﺎ ﺇﺨﻁﺎﻁ cos n θﻭ sin n θﺒﺎﺴﺘﻌﻤﺎل ﺼﻴﻐﺘﺎ ﺃﻭﻟﻴﺭ . = . sin θ = cos θﻭ ﻟﻜل θﻤﻥ ﻟﺩﻴﻨﺎ :
-VIاﻟﺠﺬور ﻣﻦ اﻟﺮﺗﺒﺔ nﻟﻌﺪد ﻋﻘﺪي ﻏﻴﺮ ﻣﻨﻌﺪم: 2i 2
1(- ﺘﻌﺭﻴﻑ: e inθ − e −inθ e inθ + e − inθ
∈ ∀n = : sin nθ = cos nθﻭ ﻭ ﺒﺼﻔﺔ ﻋﺎﻤﺔ :
ﻟﻴﻜﻥ aﻋﺩﺩﺍ ﻋﻘﺩﻴﺎ ﻏﻴﺭ ﻤﻨﻌﺩﻡ ﻭ nﻋﺩﺩ ﺼﺤﻴﺢ ﻁﺒﻴﻌﻲ ﺒﺤﻴﺙ 2 ≥ . n 2i 2
ﻜل ﻋﺩﺩ ﻋﻘﺩﻱ zﺤل ﻟﻠﻤﻌﺎﺩﻟﺔ z n = aﻴﺴﻤﻰ ﺠﺫﺭﺍ ﻤﻥ ﺍﻟﺭﺘﺒﺔ ) nﺃﻭ ﺠﺫﺭﺍ ﻨﻭﻨﻴﺎ ( • ﺘﻤﺭﻴﻥ41:
θ θ
ﻟﻠﻌﺪد اﻟﻌﻘﺪي . a θ i θ i
1 − e i θ = −2i sin e 2
1 + e i θ = 2 cos eﻭ 2
ﻟﺩﻴﻨﺎ : ﺒﻴﻥ ﺃﻨﻪ ﻟﻜل θﻤﻥ
ﻳﺴﻤﻰ ﺟﺬرا ﻣﻜﻌﺒﺎ . آﻞ ﺟﺬر ﻣﻦ اﻟﺮﺗﺒﺔ 2 ﻳﺴﻤﻰ ﺟﺬرا ﻣﺮﺑﻌﺎ ، و آﻞ ﺟﺬر ﻣﻦ اﻟﺮﺗﺒﺔ 3 2 2
-8-
![Prof : BEN ELKHATIR L’ensemble des nombres complexes ﺍﻟﺜﺎﻨﻴﺔ ﺒﺎﻜﺎﻟﻭﺭﻴﺎ ﻋﻠﻭﻡ ﺘﺠﺭﻴﺒﻴﺔ 30 ﻣﺠﻤﻮﻋﺔ اﻷﻋﺪاد اﻟﻌﻘﺪﻳﺔ
• ﺨﺎﺼﻴﺔ30: ﻟﻜل ﻋﺩﺩﻴﻥ ﻋﻘﺩﻴﻴﻥ 1 zﻭ 2 zﻟﺩﻴﻨﺎ :
ﻤﻬﻤﺎ ﺘﻜﻥ ﺍﻟﻤﺘﺠﻬﺘﺎﻥ 1 uﻭ 2 uﻭ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﺤﻘﻴﻘﻴﺎﻥ αﻭ βﻟﺩﻴﻨﺎ : ∈ αﻓﺈﻥ : 1 . α .z 1 = α .z 2 z 1 + z 2 = z 1 + zﻭ 2 z 1.z 2 = z 1.zﻭ ﺇﺫﺍ ﻜﺎﻥ
( ) ) ( ( ) ( ) ( )
2 aff u1 + u 2 = aff u1 + aff uﻭ 1aff α .u1 = α .aff u 1 ⎞ 1 ⎛
=⎟ ⎜ .
⎛z ⎞ z
ﻭ ﺇﺫﺍ ﻜﺎﻥ 0 ≠ 2 zﻓﺈﻥ : 1 = ⎟ 1 ⎜ ﻭ
(
1. aff α .u ﻭ ﺒﺼﻔﺔ ﻋﺎﻤﺔ : ) + β .u ) = α .aff (u ) + β .aff (u
2 1 2
2⎝z2 ⎠ z 2⎝z2 ⎠ z
) (
n
ﻭ ﻟﻜل ﻨﻘﻁﺘﻴﻥ Aﻭ Bﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ﺍﻟﻌﻘﺩﻱ ) ( Ρﻟﺩﻴﻨﺎ : . zn = z * *
ﻟﺩﻴﻨﺎ : ﻭ ﻟﻜل nﻤﻥ ﻭ ﺒﺼﻔﺔ ﺨﺎﺼﺔ ، ﻟﻜل zﻤﻥ
) ( ) ( ) (
) . aff AB = aff OB − aff OA = aff ( B ) − aff ( A -IIاﻟﺘﻤﺜﻴﻞ اﻟﻬﻨﺪﺳﻲ و ﻣﻌﻴﻠﺮ ﻋﺪد ﻋﻘﺪي:
• ﺨﺎﺼﻴﺔ40:
) 2 (e1 , eو اﻟﻤﺴﺘﻮى اﻟﻤﻮﺟﻪ ) ( Ρ اﻟﻤﺴﺘﻮى اﻟﻤﺘﺠﻬﻲ 2 Vﻣﻨﺴﻮب إﻟﻰ أﺳﺎس ﻣﺘﻌﺎﻣﺪ و ﻣﻤﻨﻈﻢ
ﺣﻴﺚ 0 ≠ α + β }) {( A , α ) ; ( B , β ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﻨﻘﻁﺔ Gﻤﺭﺠﺤﺎ ﻟﻠﻨﻅﻤﺔ ﺍﻟﻤﺘﺯﻨﺔ ( )
ﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ و ﻣﻤﻨﻈﻢ 2 . O , e1 , e
) α .aff ( A ) + β .aff ( B
= ) . aff (G ﻓﺈن : ( )
) M ( x , yﺗﺸﻴﺮ إﻟﻰ اﻟﻨﻘﻄﺔ Mو زوج إﺣﺪاﺛﻴﺘﻴﻬﺎ ) ( x , yﻓﻲ اﻟﻤﻌﻠﻢ 2 . O , e1 , e
α +β
1(- ﺼﻭﺭﺓ ﻋﺩﺩ ﻋﻘﺩﻱ ﻭ ﻟﺤﻕ ﻨﻘﻁﺔ ﺃﻭ ﻤﺘﺠﻬﺔ:
) aff ( A ) + aff ( B
= ) . aff ( I ﺑﺼﻔﺔ ﺧﺎﺻﺔ إذا آﺎن Iهﻮ ﻣﻨﺘﺼﻒ ] [ ABﻓﺈن : • ﺘﻌﺭﻴﻑ10:
2 ﻟﻜل ﻋﺩﺩ ﻋﻘﺩﻱ z = x + iyﺍﻟﻨﻘﻁﺔ ) M ( x , yﺗﺴﻤﻰ ﺻﻮرة اﻟﻌﺪد z
• ﻤﻠﺤﻭﻅﺔ:
ﻭ ﻋﻜﺴﻴﺎ ، ﻟﻜل ﻨﻘﻁﺔ ) M ( x , yﻣﻦ ) ( Ρاﻟﻌﺪد اﻟﻌﻘﺪي z = x + iyﻴﺴﻤﻰ ﻟﺤﻕ ﺍﻟﻨﻘﻁﺔ M
) ( ) (
ﺍﻟﺭﺒﺎﻋﻲ ﺍﻟﺫﻱ ِﺅﻭﺴﻪ ﺍﻟﻨﻘﻁ ) M ( zﻭ M zﻭ ) M ( − zﻭ M − zﻤﺴﺘﻁﻴل ﻤﺭﻜﺯﻩ
ﺭ
z = aff ( Mأو ﺑﺎﺧﺘﺼﺎر ) . M ( z ) ﻧﻜﺘﺐ :
. −( ∪i ﻤﻥ ) ﺍﻟﻨﻘﻁﺔ Oﻤﻬﻤﺎ ﻴﻜﻥ z
ﻭ ﻴﻜﻭﻥ ﻫﺫﺍ ﺍﻟﻤﺴﺘﻁﻴل ﻤﺭﺒﻌﺎ ﺇﺫﺍ ﻭ ﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ : ) . Re ( z ) = Im ( z
و )(Ρ ( )
اﻟﻤﺴﺘﻘﻴﻢ 1 O , eﻳﺴﻤﻰ اﻟﻤﺤﻮر اﻟﺤﻘﻴﻘﻲ ، 2 O , eﻳﺴﻤﻰ اﻟﻤﺤﻮر اﻟﺘﺨﻴﻠﻲ ( )
ﻳﺴﻤﻰ اﻟﻤﺴﺘﻮى اﻟﻌﻘﺪي .
2(- ﻤﻌﻴﺎﺭ ﻋﺩﺩ ﻋﻘﺩﻱ:
ﻤﻠﺤﻭﻅﺔ: •
• ﺘﻌﺭﻴﻑ:
. z ∈i ∈ zﻭ ) ⇔ M ( z ) ∈ (Oy ﻟﺩﻴﻨﺎ : ) ⇔ M ( z ) ∈ (Ox
ﻟﻴﻜﻥ z = x + iyﻋﺩﺩﺍ ﻋﻘﺩﻴﺎ ﻭ Mﻨﻘﻁﺔ ﻤﻥ ) ( Ρﻟﺤﻘﻬﺎ z
∈z − ∈ zﻭ ) ' ⇔ M ( z ) ∈ ⎡Ox
⎣ + ﻭ ﺒﺼﻔﺔ ﺨﺎﺼﺔ : ) ⇔ M ( z ) ∈ [Ox
ﺍﻟﻌﺩﺩ ﺍﻟﻤﻭﺠﺏ OM = x + yﻴﺴﻤﻰ ﻤﻌﻴﺎﺭ ﺍﻟﻌﺩﺩ ﺍﻟﻌﻘﺩﻱ zﻭ ﻴﺭﻤﺯ ﻟﻪ ﺒﺎﻟﺭﻤﺯ z 2 2
. z ∈i − z ∈ iﻭ ) ' ⇔ M ( z ) ∈ ⎡Oy
⎣ + ) ⇔ M ( z ) ∈ [Oy
. z = x +y 2 2
ﺇﺫﻥ :
• ﺘﻌﺭﻴﻑ20:
ﻤﻠﺤﻭﻅﺔ: • ﻟﻜل ﻋﺩﺩ ﻋﻘﺩﻱ z = x + iyﺍﻟﻤﺘﺠﻬﺔ ) u ( x , yﺘﺴﻤﻰ ﺼﻭﺭﺓ ﺍﻟﻌﺩﺩ ، zﻭ ﻋﻜﺴﻴﺎ ﻟﻜل ﻤﺘﺠﻬﺔ
2 ، z .z = x 2 + yﺇﺫﻥ : z = z .z ﻟﺩﻴﻨﺎ : ﻤﻬﻤﺎ ﻴﻜﻥ z = x + iyﻤﻥ
)(
) u ( x , yﻤﻥ 2 Vاﻟﻌﺪد اﻟﻌﻘﺪي z = x + iyﻴﺴﻤﻰ ﻟﺤﻕ uﻭ ﻨﻜﺘﺏ : . z = aff u
ﻭ ﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ ، ﻟﻜل ﻨﻘﻁﺘﻴﻥ Aﻭ Bﻤﻥ ) ( Ρﻟﺩﻴﻨﺎ : . AB = AB = z B − z A
. ﺇﺫﻥ ﻟﻜل ﻨﻘﻁﺔ Mﻣﻦ ) ( Ρﻟﺪﻳﻨﺎ : ) aff (OM ) = aff ( M
-2-](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)