These documents provide various math exercises on diverse subjects covered in calculus such as complex numbers, matrix, primitives, and so on. They come from a class in the top French high school Louis-Le-Grand founded by Louis XIV.
N° 8 page 17 u 0 = 1, u n+1 = (1 + i) × u n .
1. u 0 = 1 , u1 = 1 + i , u 2 = (1 + i)2 = 2i , u 3 = 2i(1 + i) = −2 + 2i .
2. (u n ) est la suite géométrique de raison q = 1 + i et de premier terme u 0 = 1.
3. u n = q n u 0 , donc u n = (1 + i)n .
¢4
4. u 8 = (1 + i)8 = (1 + i)2 = (2i)4 = 24 i4 = 16 car i4 = (i2 )2 = (−1)2 = 1 .
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Complément : Après avoir remarqué que u 4 = −4 = −4u 0 , on voit que u 5 = −4u 1 , u 6 = −4u 2 ,
u 7 = −4u 3 , etc, c’est-à-dire que les suites (u 4p ) , (u 4p+1 ) , (u 4p+2 ) , (u 4p+3 ) sont des suites géométriques
de raison −4 , ce qui permet de connaître la forme algébrique de u n de la façon suivante :
n 4p 4p + 1 4p + 2 4p + 3
p p p
un (−4) (−4) (1 + i) (−4) (2i) (−4)p (−2 + 2i)
ou si l’on préfère :
n 4p 4p + 1 4p + 2 4p + 3
p 2p p 2p p 2p+1 p 2p+1
un (−1) 2 (−1) 2 (1 + i) (−1) 2 i (−1) 2 (−1 + i)
N° 18 page 19
1. Z = z 2 − iz + 3i − 4 Z = z 2 − iz + 3i − 4 = z 2 − iz + (−4 + 3i) = z 2 − i × z + (−4 − 3i) = z 2 + i z − 3i − 4 .
2. Z = 3i + (2 + i)z Z = 3i + (2 + i) z = −3i + (2 − i)z .
µ ¶
3z + i 3z + i 3z + i 3z − i
3. Z = Z= = = .
z −i z −i z −i z +i
N° 20 page 19
On pose Z = x + yi et z = s + t i . On suppose ici que Z = z n .
On a donc x 2 + y 2 = Z Z et s 2 + t 2 = z z .
¡ ¢n ¢n
Or Z Z = z n z n = z n z n = z z , donc x 2 + y 2 = s 2 + t 2 .
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N° 21 page 19
Soit P (z) = az 2 + bz + c où a, b, c sont des réels.
Si z 0 est une racine de P (z) , on a P (z 0 ) = 0 , ce qui entraîne que P (z 0 ) = 0 = 0 .
Or P (z 0 ) = a z 0 2 + b z 0 + c , et comme a, b, c sont des réels, a = a , b = b , c = c , donc P (z 0 ) = a z 0 2 + b z 0 + c .
D’où a z 0 2 + b z 0 + c = 0 , ce qui exprime que z 0 est une racine de P (z).
Remarques :
1) Si les coefficients, ou certains coefficients, ne sont pas réels, le résultat n’est plus vrai. Par exemple, on
constate que z 0 = i est une racine du polynôme z 2 + (1 − i)z − i , mais que son conjugué z 0 = −i n’en est
pas une.
2) Le résultat se généralise facilement à un polynôme de degré n à coefficients réels :
si P (z) = a n z n + a n−1 z n−1 + · · · + a 1 z + a 0 et si a 0 , a 1 , . . . , a n sont tous réels, alors on établit comme ci-dessus
¡ ¢
que, pour tout z, P (z) = P z .
Donc z 0 est une racine de P (z) si et seulement si z 0 en est une également.
1
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