jlyp3.py

Created on February 27, 2022
1.44 KB
1.a
z1=(1+i)z0-i=-i, z2=(1+i)z1-i=(1+i)(-i)-i=-i+1-i=1-2i
z3=(1+i)(1-2i)-i=1-2i+i+2-i=3-2i

1.b
      1   2   3
 0.   oB  .   .

-1oA1 .   .   .
 
-2.   oA2 .   oA3

1.c
(z2-z1)/(zb-z1)=(1-2i+i)/(1+i)
               =(1-i)/(1+i)
               =((1-i)**2)/(1+1)
               =(1-2i+1)/2=-i
               
|(z2-z1)/(zb-z1)|=|-i|
|(z2-z1)|/|(zb-z1)|=1
|(z2-z1)|=|(zb-z1)|
A1A2=A1B

(>A1A2;>A1B)=arg(z2-z1)/(zb-z1)=arg(-i)=-pi/2

Le triangle BA1A2 est isocele recangle en A1

2.a
un+1=|zn+1-1|=|(1+i)zn-i-1|
=|(1+i)(zn-1)|=|1+i|*|zn-1|
=racine(2un)

un+1/un=racine(2), la suite (un)
est géométrique de raison q=racine(2)
et de premier terme u0=|-1|=1

2.b
On a alors un=u0*q**n=(racine(2))**n
On veut donc (racine(2))**n > 1000
Sachant que 2**10=(2)**20=1024,
on en déduit que n=20

3.a
1+i=racine(2)*((racine(2)/2)+(racine(2)/2)*i)
racine(2)*e**(i*pi/4)

3.b
Initialisation: n=0, on a 
1-(racine(2)**0*e**0)=1-1=0=z0

Hérédité : Soit n appartient N,
supposons que zn=1-((racine(2))**n)*e**(i*pi/4)
zn+1=(1+i)zn-i=(1+i)(1-(racine(2)**n)*e**(i*n*pi/4))-i
=1+i-(1+i)(racine(2)**n)*e**(i*n*pi/4)-i
=1-racine(2)*e**(i*pi/4)*(racine(2)**n)*e**(i*n*pi/4)
=1-(racine(2)**n+1)*e**(i*(n+1)*pi/4)
La propriete est hereditaire

Conclusion: zn=1-(racine(2)**n)*e**(i*n*pi/4)

3.c
z2020=1-(racine(2)**2020)*e**(i*2020*pi/4)
     =1-(racine(2)**2020)e**(i*505*pi)
     =1-(racine(2)**2020)*e**i*pi
     =1+(racine(2)**2020)
Le point A2020 est donc sur l'axe des abscisses