Bonjour,
[tex]1)\\
z_0= \sqrt{3}-i\\
2)\\
u_0^2= (\sqrt{3})^2+1^2=4+1=4\\
u_0=2\\
z_{n+1}= z_{n}*(1+i)\\
z_1= (\sqrt{3}-i)*(1+i)=1+ \sqrt{3} -i+i \sqrt{3} \\
u_1^2= (1+ \sqrt{3})^2+(-1+\sqrt{3})^2 =1+3+2*1* \sqrt{3}+1+3-2*1* \sqrt{3} \\
=8\\
u_1=2 \sqrt{2} =u_0* \sqrt{3} \\
[/tex]
[tex]3)\\
u_1^2= 2*u_0^2\\
z_n=a+i*b\\
u_n^2=a^2+b^2\\
==\textgreater\ z_{n+1}=z_n*(1+i)=(a-b)+i(a+b)\\
u_{n+1}^2=(a^2+b^2+2ab)+(a^2+b^2-2ab)=2(a^2+b^2)=2*u_n^2\\
[/tex]
[tex]4)\\
u_0^2=2\\
u_1^2=2^1*2\\
u_2^2=2^2*2\\
...\\
u_n^2=2^n*2\\\\
\boxed{u_n=2^{ \frac{n}{2} +1}} \\
\lim_{n \to \infty} u_n =\infty\\
[/tex]
5)
Traitement:
Tant que u<p
Affecter n+1 à n
Affecter √2*u àu
fin tantque
Sortie:
afficher n,p,u
fin
p=-10==>n=0
p=1==>n=0
p=10==>n=6
p=200==>n=15
p=5000==>n=24
Voici le calcul pour 10^6
[tex]2^{ \frac{n}{2}+1}\ \textgreater\ 10^6\\
( \frac{n}{2}+1)*ln(2)\ \textgreater \ 6*ln(10)\\
\frac{n}{2}+1\ \textgreater \ 6* \frac{ln(10)}{ln(2)} \\
\frac{n}{2}\ \textgreater \ (6* \frac{ln(10)}{ln(2)}-1) \\
n \ \textgreater \ 38
[/tex]
