Articles
Register
Sign In
Search
alcath08
@alcath08
January 2021
1
131
Report
bonsoir,niveau terminale s
quelqu'un peut m'aider svp merci
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms and service
You must agree before submitting.
Send
Lista de comentários
taalbabachir
Verified answer
A) z² = - 1 on sait que i² = - 1
z² = i² ⇔ z² - i² = 0 ⇔ (z - i)(z + i) = 0 ⇒ z = i ou z = - i
b) z² - 4 z + 13 = 0
Δ = 16 - 4*13 = 16 - 52 = - 36 on pose i² = - 1 ; √(6i)² = 6i
z1 = (4 + 6i)/2 = 2 + 3i
z2 = (4 - 6i)/2 = 2 - 3i
c) z - 3i = - 2iz + 4 ⇔ z + 2iz = 4 + 3i ⇔ z(1 + 2i) = 4 + 3i ⇒ z = (4 + 3i)/(1+2i)
⇒ z = (4 + 3i)/(1+2i) ⇔ (4 + 3i)(1 - 2i)/(1 + 2i)(1 - 2i) = (4 - 8i + 3i - 6i²)/(1-4i²
⇒ z = (10 - 5i)/5 = 2 - i
vous faite la suite
1 votes
Thanks 1
alcath08
merci beaucoup
More Questions From This User
See All
alcath08
June 2021 | 0 Respostas
Responda
alcath08
June 2021 | 0 Respostas
Responda
alcath08
June 2021 | 0 Respostas
bonjour ,pouvez-vous me traduire : etes vous content d'etre president. merci
Responda
alcath08
February 2021 | 0 Respostas
Responda
alcath08
February 2021 | 0 Respostas
Responda
alcath08
February 2021 | 0 Respostas
Responda
alcath08
February 2021 | 0 Respostas
Responda
alcath08
January 2021 | 0 Respostas
Responda
alcath08
January 2021 | 0 Respostas
Responda
alcath08
January 2021 | 0 Respostas
Responda
×
Report "bonsoir,niveau terminale s quelqu'un peut m'aider svp merci.... Pergunta de ideia de alcath08"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
Sobre nós
Política de Privacidade
Termos e Condições
direito autoral
Contate-Nos
Helpful Social
Get monthly updates
Submit
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
Verified answer
A) z² = - 1 on sait que i² = - 1z² = i² ⇔ z² - i² = 0 ⇔ (z - i)(z + i) = 0 ⇒ z = i ou z = - i
b) z² - 4 z + 13 = 0
Δ = 16 - 4*13 = 16 - 52 = - 36 on pose i² = - 1 ; √(6i)² = 6i
z1 = (4 + 6i)/2 = 2 + 3i
z2 = (4 - 6i)/2 = 2 - 3i
c) z - 3i = - 2iz + 4 ⇔ z + 2iz = 4 + 3i ⇔ z(1 + 2i) = 4 + 3i ⇒ z = (4 + 3i)/(1+2i)
⇒ z = (4 + 3i)/(1+2i) ⇔ (4 + 3i)(1 - 2i)/(1 + 2i)(1 - 2i) = (4 - 8i + 3i - 6i²)/(1-4i²
⇒ z = (10 - 5i)/5 = 2 - i
vous faite la suite