QCM corrigé nombres complexes 0 analyse numérique, arithmétique, corrigé, denombrement, géométrie analytique, mathematique, nombres complexes, probabilité, QCM, series, suites 09 h 14 QCM nombres complexes corrigé math analyse numérique licence QCM Nombres complexes Q u e s t i o n 1 Le nombre complexe 1 + i 3 1 − i 20 v a u t : □ 2 9 □ i 2 9 □ 2 9 1 + i 3 ☑ 2 9 1 − i 3 Q u e s t i o n 2 Soit u racine 5 i è me de l ' u n i t é et u ≠ 1 , alors 1 + u + u 2 + u 3 est é gale : □ 1 u ☑ − 1 u □ 1 u − 1 □ 1 u + 1 Q u e s t i o n 3 ∑ k = 1 10 sin 2 k π 7 + icos 2 k π 7 est é g a l e : □ 1 □ − 1 □ i ☑ − i Q u e s t i o n 4 le module du nombre complexe Z = 2 + i 5 2 − i 5 10 + 2 − i 5 2 + i 5 10 v a u t : ☑ 2 c o s 20 arcos 2 3 □ 2 s i n 20 arcos 2 3 □ 2 c o s 10 arcos 2 3 □ 20 cos 2 arcos 2 3 Q u e s t i o n 5 la partie r é elle du nombre comple ω = 1 − cos θ + 2 i s i n θ − 1 est é g a l e : □ 1 3 + 5 c o s θ ☑ 1 5 + 3 c o s θ □ 1 3 − 5 c o s θ □ 1 5 − 3 c o s θ Q u e s t i o n 6 Soit ω le point d ' a f f i x e i . L ' e x p r e s s i o n complexe du rotation de centre ω et d ' a n g l e π 2 est ☑ z ' = iz + 1 + i □ z ' = − iz + 1 + i □ z ' = iz − 1 + i □ z ' = iz + 1 − i Q u e s t i o n 7 On considere le nombre complexe ω = 2 3 e 2 i π 3 , on pose α n = ω n + 1 − ω n Alors on a : ☑ α n = 19 3 2 3 n □ α n = 2 3 n + 1 − 2 3 n □ α n = 15 13 2 3 n □ α n = 13 15 2 3 n □ α n = 3 19 2 3 n Question 8 Soit I le po int d ' affixe 3 i . L ' ensemble des po int s d ' affixes z telque iz + 3 z − 3 i = 1 est : □ l ' axe des abscisse . □ l ' axe des ordonn é es priv é du po int I . □ une cercle de rayon 1 priv é e du po int I . ☑ le plan priv é du po int I . Question 9 On pose S = 1 + 2 ω 0 + 3 ω 2 0 + ... + n ω 0 n − 1 avec ω 0 est la racine ni è me de l ' unit é . S vaut alors : □ S = 1 ω 0 − n □ S = n ω 0 − n □ S = 1 ω 0 − 1 ☑ S = n ω 0 − 1 Question 10 − 1 + − 3 2 3 n + − 1 − − 3 2 3 n vaut : □ − 1 □ 1 ☑ 2 □ 3 MMF.7h_3?`00kEY]Lm|f4_h5oPojIVZ6dA00Go==E]GDLkJD|mBdLoEmH2C6ahU(^YBDf|kX_al00^B212QI]Zo^CN)aPmg5bnj3gLD2d_Gii(OYeNE`OSjMW7`o^AiMSHOcjIEe6FonLaWOY8_Nab;_oI9V_Fo18)ROG(o?iaMSjd^n;BSaJG`eX`(]cnkiM|nQW1n6lk55fhSnHPMAPH?OHoNmPjQ`o)_7Rn7ia)ZOk4j^In?io7cbPBdjVdf6Uf?[agRabH^7gRcIK=;|QPiHKbc7?R_RiBYij?dDIl_NnGaD;TA5b;jl72GIIU|lC98o:SJfco;lJfnHKM;O]g759WG_jF[iXDRB[iG(kI1inUEl?C_*;akJ8hYHVQ*mf3VbinU]|VJlgUEn6fNeJHin7VB`6F43_l]Ze6Df(]R=38HSPnE8KcXbfhh=]V?COQ]|aeffhbkK|FO*`60k;VfWA_CXG;gIiV6Ee48m0=P(077dba0304*;`2(=7F`SE;JASNa*l2WGm_fbCFPKnEi9^:aCF;Hm9|12h5(R4)f0]SdaDlQ72b82Be3oAGHin|iRKBCJ2;*aj(?dl0IQBKQ0`?*8gK;]0gi0SG(6A50QYd*gYPQba7Y)YNhMRaiTbcIVj8PfFmfG6[;EPh6Tf?ZbkL)96*j8B8XQhD|RDWC33QR6P*ZH0A19PT1TG4QhH0QD0*NeKSQDeh*H4;hIF10804;h5TR2E1]82KP7a5(V9gjm;PV0hBB4LdN0L246;WM6(H6;PJT^PMeLB9*Z^;8SDn5M9AC)8=dW18Jh4G*Vj8XNE(=SJXSa7]`:cfEn8]g1hf4Q[O9lj0(Ndd:Pk8FP7MFNiS^PSFXHO*cJ40BO;bl9[ikEml5(K1)8e((?^LFBRV[_2F0`1=0A0[1n`=HWX^ebI0=1NG08(ej^6*1G3=Sj8Q*3kXM2iI0]ClAL8E`n;?e*DPB:X2)6YAM8bZm=2iT2kjYd4099E:|F)J0=8S72H(F8e(Q6;W3_b0(K4oU`B00l9PXQ4H48*hjSi1H7ZBAF4QL1VLMa8N7E6h*L9BTiPD:542kT04M4b55TBVY4?35E8|:Q[HAUF5JT1`OjT0Q|]bHH:)nZ:AEDN8:|MQoa58TUPB718d(B41:N8g55`Ke1?4f65*EL0f4*78RWbHY0((4P0^83dDCYe1eMT5bAb9ANeEGI7Ih|:b1i]ZbY2Y8?`hodb7A|[[I=j?lUEJk9Y8QCZYCn22Wg8Rj7ob(Q9GAD:?RQn1_bc(NT;YfA20VgT]=4B3g)JOd8ZLne(6TEE=:0B`=56WJ)SK[Af0(FHU_BnV5hG4fWLl|Id7Qb1SkmAJa1WHbbn;U`nlEBBk6BQc*lg2hWB`7AM=KDWZF0eMc3X|Q5?eI^WZEIG3bDM43YBG8CKi:B3RUmUJcRCIYW9BNRW4maTLJOE`TFfS8CaY^5X9Sbho]=4*|JPa53`B)0MbIh;^2=1(l3_1l4cmL15^RHX*T55)T0MTcMnKd0X(K_0a0fOQ60^?7RG`D)NfgTN;UO8HL35CTL]Y73DA|iG|dgT))EO0)iAQD_V3[g8DKo8Ij^^fo|7ZS8TK2178TJb;U)2cTG]I5c(DC)9BYb[]]6c_GJb;Un6cTgJ20G[fSD|:A)ceIJH=0BRMF=9N7H[)idF8?b2K_ba8GBWC1fV*lLgU]bGEK0S2j^J0:Hi(E]_)[=To^=b8TRP`c:;(9j|F?Uj|?I^k?E=Q5YUjgY4*mfW5i(Va?BE(?oJAYdF3E8VEoK5Hhin`OGW;IV?emJ;7^OCnHFDf7fdoBG|n5|c8*k=_UT?Yk(fA_(9C]cn]n_ckiOSgjcKM^gfCnfk[n]Ohh^nmLcV_okKMTT_oeL96^SO97OgZfBnj[7C[o2=UTcUd8])AEN99[I:CoSBn]ULUTYYGVN7V?L_;WU=na4kP3cDHQfojfLX^;Q2?DeGM)6YYecT23Dcd6E*JR5Ncf);_:Y)IWCgfVhV70f7D;o^7*LWCUXkBO[b(O[(?|FKcO_Bk=Vm=KRj975ON?D4Ieigo3dVN?g38LK[D[I_S*fXW(CjPd*R5O*2QGmb7d=5EEoNhj:aRS3^RRKiNU6KlmFcbkRAIXI`|nS:VmED0Q^M[Xe35hJn:_CKLIdK4i(m=fC3W?XZN`fnNSJePn8EgVQBg0MC[DelheabL?bRH)8:IKYAT^==G2YOFoRU*5aGCKXB9mKLdhdIZaV:8T0N8eE3SGRh3)PNE[m3c*aW2A7JV;(4:BLKlhZ*6gFFJOIWT)EM`]dV*Io;J?3iiV[NG1Y1_3n:0cJdJOYC6oK^n|;FU2bX||*DNTRGknBnRlBSO]?HJCYXCdkM1fKYh|a2[DMSDkSj]0bYIKKO;TebIJ64nLY1IlZoEOU*HLEP_PYQN2N8|hc5W47;ge|oFQNn_Sj4CWPX3XNcILgjDoLb6NS^JlJij7IB4?=7=HZloG5C75PkR?Jg6M8fJgRf5bo=dn)_hSJgFRS]jZfPKo7Moi4HhbWBo_YP9d^|EkE^kSHY8JcX3SP44iF;glfeFL[L*?MNGCHII3_`LabOLM`al0Y3k6bRb49VDZ5eb]O)Xh7L_Sah9GF=g^GUSkgj]6QXWNhR^C5EC`NAL=9mo8X7ZJR3TF3RRn=XS6]n?_N*VBhjQhT3?5nUjNIhO5QNAYonI:Jb]AdX7go?4g^khYT_FI_kTLD_jJDENBKlQgoBJl`RbCK5:eTmXCg5`Y1M[=:)_`53d9K7QW(_NB^B[jhNcGIK;)=nF_OXmgSZN6BVChJlZci*G7OZEb]IN]^FJnhWWh]kE7`BV^Ik3[R^ZVmW4n=4I:]dfER)Z5=TObD6fOcG(Nn;99PiLo(*0=R=n|Tb(O*^b|Vjlfk1SK8L8UlZFT|4PdR[5^Th|_WF(Q;jiAKSeN2[Ai)n6?mb6:_YRX`YXabUjl1F1h72FT;b:blHSH;986`jBjE6Af(clIMK)lD)ejBH]`_onkdWe[A[T?cZk0904)anUjCW9j2CdM987bSLJ^5OE3d]5_]PEOniQD0G(S=efG1MZRjXZ9Y|[Z^dYWfdj)S7)f]6KKO[mn`l|oMUMYi^cji=IZ6](lfZVg*kon6|YW^E93oQ_89D4HVdokoT9*WBk=(jkLTXM1BoLSI)ZnKX1bSE*?k2bmDeY;m)FEo6YLYY_b8EC4SBmjT[8^ZLi:]Tm_?L)hUn)8:G6QT[Lf[0M5SgM`TZmngnkm^0_5ok;`)Zo_Ji|7bF)7[YZe?Sd)MD2V1A3g?BgEQ;gPC6VTo9jJkF0=gK`8mo[aNY6]F5eIKM|*|NK7((nYYFo4RIBWkFZCO:U4eO=_YHNMVOKKP)`od(U`|E8NZfdGl00Y1R9fR4EGfB;GPGEH=VEFL_IcmQ`Nfc7K`5ZIDoCCTnjXg2LoTCliEJBh3aX8O=O7PKYO_iOLVc8(6P`5H9S(_0nO(jR]YFFEE_N|GDR_nUR`JMknf6[YX0HoFcBoAF5UJKQFkWL4_b2RNEHN4n4:(m:*ZjLgZc^ac911_iJg55;;emUAYZie;g?HQfCR^MXLV]|b*a9k`g_QfU(kN|];XB:BE9?=FU)i6n_24QAai*nhh[Sg`:P(=[_R:jY8;_`HZNN8SlkYd?70^iIbFC?F:kX5GXV=Z^[9bb^ZL9E;V0NG(bn4SK[9o2GaTH]EWDWQjJ88OJHm`[)eKe_C:kniTMo9OEXFKM`.mmf TELECHARGER VERSION PDF QCM corrigé nombres complexes QCM nombres complexes corrigé math analyse numérique licence QCM ... 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QCM Suites numériques 0 analyse numérique, arithmétique, corrigé, mathematique, QCM, QCM analyse numérique, QCM calcul de somme, QCM convergence, QCM développement limité, qcm suites numeriques, séries, suites 17 h 44 MathMagic Equation - www.mathmagic.com QCM Suites n u m é r i q u e s Q u e s t i o n 1 : la suite de terme g é n é r a l u n n ∊ N * = cos 1 n n 2 tend vers : □ + ∞ □ e □ 1 2 e ☑ e − 1 2 □ n ' a d m e t pas de limite Q u e s t i o n 2 : Soit la suite u n n ∊ N * = 1 + 1 2 + 1 3 + ... + 1 n □ la suite u n est convergente □ la suite u n est positive □ la suite u n est m a j o r é e ☑ la suite u n est croissante Q u e s t i o n 3 : La somme S = ∑ n = 1 + ∞ n 5 n vaut : □ 1 4 □ 4 5 ☑ 5 16 □ 1 16 Q u e s t i o n 4 : Soit la suite de terme g é n é r a l a n = a 2 n − 1 + a 2 n + 1 2 , ∀ n ≥ 1 ☑ la suite a n est constante ou divergente □ la suite a n est arithmetique □ la suite a n est g é o m e t r i q u e □ On peut pas en conclure Q u e s t i o n 5 : La suite x n n ∊ N definit par x n = π − n − π 10 est : □ b o r n é e □ croissante □ convergente ☑ d é c r o i s s a n t e Q u e s t i o n 6 On note S n = v 0 + v 1 + ... + v n pour tout n ∊ N avec v n = ln n + 2 n + 3 Alors : □ lim n → + ∞ S n = 0 □ lim n → + ∞ S n = 1 □ lim n → + ∞ S n = 2 □ lim n → + ∞ S n = 3 ☑ lim n → + ∞ S n = − ∞ Q u e s t i o n 7 Soit la suite α n d é f i n i e par α n = n − n + a n + b avec a , b ∊ R On a alors : □ lim n → + ∞ α n = a − b □ lim n → + ∞ α n = − a + b ☑ lim n → + ∞ α n = − a + b 2 □ lim n → + ∞ α n = a + b Q u e s t i o n 8 ∑ k = 1 n − 2 n − k C n k vaut : □ 1 − − 1 n □ 1 − − 2 n □ 0 ☑ − 1 n − − 2 n □ − 1 n + − 2 n 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