Viens dans C

C
F
e
x2=−1.
C
e

C
x3+px =q
x=3
sq
2−rq2 4+p3 27 +3
sq
2+rq2 4+p3 27
q2
4+p3 27 ≥0.
x3−3x= 2 p q
x3−15x= 4
−1√−1
√−1
−1 = (√−1)2=√−1√−1 = p(−1)2= 1 !
√−1
i√−1
i2=−1
(2 + i)3= 2 + 11i(2 −i)3= 2 −11i.
x= 4
x3−15x= 4 x= 4
x3−15x= 4
x3−15x= 4
x2+ 1 = 0
i−i
C
R
•iCi2=−1
•Ca+ib a b
BY:
$
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C

C
•
z
•z=a+ib a <e(z)
•z=a+ib b =m(z)
•z=a+ib √a2+b2|z|
•z=a+ib a −ib z
z1= 2 −5i z2= 1 + 3i a +ib a b
z1+z2z1z2
z= 2 + 3i
z2|z|2
zz
z1z2z1=i
4−3iz2=1+i 1−i
a+ib a b z1z21
z1
1
z2
z1z2
|z1| |z2|
z1z2z1z2=z1z2
|z1||z2|=|z1z2|
p q p q
a, b, c d
p=a2+b2q=c2+d2.
pq
(O, −→ i , −→ j)
•z=a+ib M(z)
a b −−→ OM =a−→ i+b−→ j
BY:
$
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C
•M a b
z=a+ib M
01i1 + i
z
|z|
M N P Q z z −z3z
N P Q M
M z
|z|=|z−1|=1
|z|
|z−i|=|z+ 1| |z+i|=|z−2|
z θ
[0,2π[z z =|z|(cos θ+isin θ)
z
eiθ = cos θ+isin θ
z=|z|eiθ
eiπ
2eiπ 4eiπ e2iπ
θ1θ2ei(θ1+θ2)=eiθ1eiθ2
z=|z|eiθ
iz
M
z M0iz
z=1
√3+i
3
z
z7
z1=eiπ
3z2=e−iπ 4
a+ib a b z1z2z1z2
cos( π
12 ) sin( π 12 )
n x
S1=
n
X
k=0
cos(kx)S2=
n
X
k=0
sin(kx).
BY:
$
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C

C
i2=−1
a2+b2=|a+ib|2
S1+iS2
1−eix =−2ieix/2sin(x/2).
BY:
$
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C