Contrôle de mathématiques Exercice 1 (cours) d

R∈R+Ω(ω)
z0=f(z)
−→
u(1 + 2i) Ω(2 −2i)α= 3π/4
Ω(2i)k= 3
z0= 3z+ 2i z0=z−3 + i z0=−iz + 3 z0= −1
2+i√3 2!z
z z =−z
z z =z
z zz =|z|2
(O, −→ u , −→ v)
A B C
a b c O
H a +b+c
a= 3 + i b =−1+3i c =−√5−i√5
O ABC
A B C H a +b+c
H ABC
ABC O a b c
A B C
O ABC
aa =bb =cc.
w=bc −bc
w
(b+c)b−c=wb+c
b−c=w |b−c|2
b+c
b−c
a+b+c
a, b c −−→ AH −−→ CB
−−→
CB, −−→ AH=π
2+kπ k
−→
CA, −−→ BH=π
2+kπ