Tiêu đề Complex Number Engg Practice Sheet Solution.pdf ✅

2  Higher Math 2nd Paper Chapter-3 7| 3 a + ib = x + iy n‡j cÖgvY Ki †h, 4(x2 – y 2 ) = a x + b y [BUET 11-12] mgvavb: 3 a + ib = x + iy  a + ib = x3 – 3xy2 + i (3x2 y – y 3 )  a = x3 – 3xy2 ; b = 3x2 y – y 3 a x = x 2 – 3y2 ; b y = 3x2 – y 2  a x + b y = x 2 – 3y2 + 3x2 – y 2 = 4(x2 – y 2 )  4(x2 – y 2 ) = a x + b y (Proved) 8| hw` 3 a + ib = x + iy nq Z‡e †`LvI †h, 3 a – ib = x – iy [BUET 04-05; RUET 03-04, 07-08; BUTex 03-04] mgvavb: 3 a + ib = x + iy  a + ib = x3 – 3xy2 + i (3x2 y – y 3 )  a = x3 – 3xy2 ; b = 3x2 y – y 3  a – ib = x3 – 3xy2 – 3ix2 y + iy3  a – ib = x3 – 3x2 iy + 3x(iy)2 – (iy)3  a – ib = (x – iy)3  3 a – ib = x – iy (Showed) 9| hw` x = 2 + –3 nq, Z‡e 3x4 – 17x3 + 41x2 – 35x + 5 Gi gvb wbY©q Ki| [BUET 01-02] mgvavb: x = 2 + –3  x = 2 + i 3  x – 2 = i 3  (x – 2)2 = 3i2  x 2 – 4x + 4 = –3  x 2 – 4x + 7 = 0 GLb, 3x4 – 17x3 + 41x2 – 35x + 5 = (3x4 – 12x3 + 21x2 ) – 5x3 + 20x2 – 35x + 5 = 3x2 (x2 – 4x + 7) – 5x(x2 – 4x + 7) + 5 = 0 – 0 + 5 = 5 (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 10| 3 a + ib = x + iy n‡j cÖgvY Ki †h, 4(x2 – y 2 ) = a x + b y [KUET 03-04; BUET 11-12; BUTex 04-05; RUET 08-09, 05-06] mgvavb: †`Iqv Av‡Q, 3 a + ib = x + iy  a + ib = (x + iy)3 = x 3 – iy3 + 3ix2 y – 3xy2  a = x3 – 3xy2 , b = 3x2 y – y 3  a x = x 2 – 3y2 ; b y = 3x2 – y 2  a x + b y = x 2 – 3y2 + 3x2 – y 2 = 4(x2 – y 2 )  4 (x2 – y 2 ) = a x + b y (Proved) 11| 6 – 64 Gi gvb wbY©q Ki| [KUET 03-04; CUET 15-16] mgvavb: x = 6 – 64  x 6 = – 64  (x2 ) 3 = (– 4)3      x 2 – 4 3 = 1  x 2 – 4 = 1, ,  2 x 2 = – 4, – 4, – 4 2 x =  – 4 ,  – 4  ,  – 4  2 =  2i,  2i  ,  2i (Ans.) weMZ mv‡j RUET-G Avmv cÖkœvejx 12| hw` z = cos + i sin, †`LvI †h, 2 1 + z = 1 – i tan  2 | [RUET 19-20; MIST 15-16] mgvavb: z = cos + i sin 2 1 + z = 2 1 + cos + i sin = 2 (1 + cos) + i sin × (1 + cos) – i sin (1 + cos) – i sin = 2(1 + cos) – 2i sin (1 + cos) 2 – (i sin) 2 = 2(1 + cos) – 2i sin 1 + 2cos + cos2  + sin2  = 2(1 + cos) – 2i sin 1 + 2cos + 1 = 2(1 + cos) 2(1 + cos) – 2i sin 2(1 + cos) = 1 – i × 2sin  2 cos  2 2 cos2  2 = 1 – i tan  2 13| 4  + 3 = 8–1 n‡j  KZ? [RUET 17-18] mgvavb: 4  + 3 = 8–1  2 2+6 = 23–3  2 + 6 = 3 – 3   = 9 (Ans.) 14| hw` 3 a – ib = x – iy nq, Z‡e †`LvI †h, 3 a + ib = x + iy [RUET 17-18; CUET 07-08; MIST 16-17] mgvavb: †`Iqv Av‡Q, 3 a – ib = x – iy  a – ib = x3 – 3xy2 – i (3x2 y – y 3 )  a = x3 – 3xy2 ; b = 3x2 y – y 3  a + ib = x 3 – 3xy2 + i (3x2 y – y 3 ) = x 3 + 3x (iy)2 + 3x2 (iy) + (iy)3 = (x + iy)3  3 a + ib = x + iy
RwUj msL ̈v  Engineering Practice Sheet Solution 3 15| eM©g~j wbY©q Ki: cos + i sin, i = – 1 [RUET 15-16] mgvavb: cos + i sin = e i  eM©g~j =  (ei ) 1 2 =  e i 2 =      cos  2 + i sin  2 (Ans.) 16| (a) 4 + 3i RwUj msL ̈vi gWzjvm I Av ̧©‡g›U wbY©q Ki| [RUET 12-13, 04-05] mgvavb: gWzjvm, r = 4 2 + 32 = 5  tan = 3 4  Av ̧©‡g›U,  = tan–1     3 4 (Ans.) (b) r (1 – r) = 1 Gi RwUj g~jØq z1 I z2 n‡j cÖgvY Ki †h, z1 3 + z2 3 = – 2 mgvavb: r (1– r) = 1  r 2 – r + 1 = 0  z1  z2 = 1 ... (i)  z1 + z2 = 1 ... (ii) (ii) bs n‡Z (z1 + z2) 3 = (1)3  z1 3 + z2 3 + 3z1z2 (z1 + z2) = 1  z1 3 + z2 3 + 3 = 1  z1 3 + z2 3 = 1 – 3  z1 3 + z2 3 = – 2 (Ans.) 17| gWzjvm I Av ̧©‡g›U wbY©q Ki: – 3 + i [RUET 11-12] mgvavb: gWzjvm, r = (– 3) 2 + 12 = 2 Av ̧©‡g›U,  =  – tan–1     1 – 3 =  –  6 = 5 6 (Ans.) 18| eM©g~j wbY©q Ki: 2 + i a 2 – 4 [RUET 11-12] mgvavb: awi, 2 + i a 2 – 4 =  (x + iy) GLv‡b, r = 4 + a2 – 4 = a  x 2 = a + 2 2  x = a + 2 2  y 2 = a – 2 2  y = a – 2 2  wb‡Y©q eM©g~j =       a + 2  2 + i a – 2 2 =  1 2 ( a + 2 + i a – 2) (Ans.) 19| 3 a + ib = x + iy n‡j cÖgvY Ki †h, 4(x2 – y 2 ) = a x + b y [RUET 08-09, 05-06; BUET 11-12; BUTex 04-05; KUET 03-04] mgvavb: †`Iqv Av‡Q, 3 a + ib = x + iy  a + ib = (x + iy)3 = x 3 – iy3 + 3ix2 y – 3xy2  a = x3 – 3xy2 , b = 3x2 y – y 3  a x = x 2 – 3y2 ; b y = 3x2 – y 2  a x + b y = x 2 – 3y2 + 3x2 – y 2 = 4(x2 – y 2 )  4 (x2 – y 2 ) = a x + b y (Proved) 20| wb‡Pi RwUj msL ̈vwUi gWzjvm I Av ̧©‡g›U wbY©q Ki? 1 + 3i [RUET 06-07] mgvavb: gWzjvm, r = (1) 2 + ( 3) 2 = 2 Av ̧©‡g›U,  = tan–1 3 =  3 (Ans.) 21| gvb wbY©q Ki: 3 i + 3 – i [RUET 05-06] mgvavb: x = 3 i + 3 –i x 3 = ( ) 3 i 3 + ( ) 3 – i 3 + 33 i . 3 – i( ) 3 i + 3 – i = i – i + 33 – i 2 .x = 3.1.x = 3x  x 3 = 3x  x = 0,  3 (Ans.) 22| hw` 3 a + ib = x + iy nq Z‡e †`LvI †h, 3 a – ib = x – iy [RUET 03-04, 07-08; BUET 04-05; BUTex 03-04] mgvavb: 3 a + ib = x + iy  a + ib = x3 – 3xy2 + i (3x2 y – y 3 )  a = x3 – 3xy2 ; b = 3x2 y – y 3  a – ib = x3 – 3xy2 – 3ix2 y + iy3  a – ib = x3 – 3x2 iy + 3x(iy)2 – (iy)3  a – ib = (x – iy)3  3 a – ib = x – iy (Showed) 23| (a) gWzjvm I Av ̧©‡g›U wbY©q Ki: – 4 – 3i [RUET 03-04] mgvavb: gWzjvm, r = (– 4) 2 + (– 3) 2 = 5 Av ̧©‡g›U,  = –  + tan–1 3 4 [gyL ̈ Av ̧©‡g›U] (Ans.) (b) gvb wbY©q Ki: i + i + i + .......  mgvavb: awi, x = i + i + i + .......   x 2 = i + i + i + .......   x 2 = i + x  x 2 – x – i = 0  x = 1  1 + 4i 2 (Ans.)
4  Higher Math 2nd Paper Chapter-3 weMZ mv‡j CUET-G Avmv cÖkœvejx 24| 6 – 64 Gi gvb wbY©q Ki| [CUET 15-16; KUET 03-04] mgvavb: x = 6 – 64  x 6 = – 64  (x2 ) 3 = (– 4)3      x 2 – 4 3 = 1  x 2 – 4 = 1, ,  2 x 2 = – 4, – 4, – 4 2 x =  – 4 ,  – 4  ,  – 4  2 =  2i,  2i  ,  2i (Ans.) 25| hw` 3 a – ib = x – iy nq, Z‡e †`LvI †h, 3 a + ib = x + iy [CUET 07-08; MIST 16-17; RUET 17-18] mgvavb: †`Iqv Av‡Q, 3 a – ib = x – iy  a – ib = x3 – 3xy2 – i (3x2 y – y 3 )  a = x3 – 3xy2 ; b = 3x2 y – y 3  a + ib = x 3 – 3xy2 + i (3x2 y – y 3 ) = x 3 + 3x(iy)2 + 3x2 (iy) + (iy)3 = (x + iy)3  3 a + ib = x + iy weMZ mv‡j MIST-G Avmv cÖkœvejx 26| cÖgvY Ki †h, GK‡Ki RwUj Nbg~‡ji ̧Ydj 1 [MIST 22-23] mgvavb: g‡b Kwi, 3 1 = x  1 = x3  x 3 – 1 = 0  (x – 1) (x2 + x + 1) = 0  x – 1 = 0  x = 1 Avevi, x 2 + x + 1 = 0  x = – 1  i 3 2  GK‡Ki Nbg~j ̧‡jv 1, – 1 + i 3 2 , – 1 – i 3 2 = 1, ,  2    – 1 + i 3 2    – 1 – i 3 2 = (– 1) 2 – (i 3) 2 2 2 = 1 + 3 4 = 1 RwUj Nbg~‡ji ̧Ydj = . 2 =  3 = 1 (Proved) 27| GK‡Ki Nbg~j wbY©q Ki| [MIST 20-21] mgvavb: awi, 3 1 = x  x 3 – 1 = 0  (x – 1) (x2 + x + 1) = 0  x – 1 = 0  x = 1 Avevi, x 2 + x + 1 = 0  x = – 1  1 – 4 2 = – 1  i 3 2  GK‡Ki Nbg~j ̧‡jv 1, – 1 + i 3 2 , – 1 – i 3 2 (Ans.) 28| z + i z + 2 we›`yi mÂvic‡_i mgxKiY wbY©q Ki, hLb GwU m¤ú~Y© KvíwbK| [MIST 19-20; BUET 19-20] mgvavb: z + i z + 2 = x + iy + i x + iy + 2 = x + i(y + 1) (x + 2) + iy = {x + i(y + 1)} (x + 2 – iy) (x + 2 + iy) (x + 2 – iy) = x(x + 2 – iy) + iy(x + 2 – iy) + i(x + 2 – iy) (x + 2 + iy) (x + 2 – iy) = x 2 + 2x – ixy + ixy + 2iy + y2 + ix + 2i + y (x + 2) 2 + y2 = x 2 + 2x + y2 + y + i(2y + x + 2) (x + 2) 2 + y2 m¤ú~Y© KvíwbK n‡j, x 2 + 2x + y2 + y (x + 2) 2 + y2 = 0  x 2 + y2 + 2x + y = 0 myZivs, GUv wb‡Y©q mÂvic‡_i mgxKiY, hv GKwU e„Ë wb‡`©k K‡i| 29| – i Gi Nbg~j wZbwUi †hvMdj wbY©q Ki| [MIST 18-19] mgvavb: x 3 = – i      x i 3 = 1  x i = 1, ,  2  x = i, i, i 2  Nbg~j wZbwUi †hvMdj: i (1 +  +  2 ) = 0 30| hw` 3 a – ib = x – iy nq, Z‡e †`LvI †h, 3 a + ib = x + iy| [MIST 16-17; RUET 17-18; CUET 07-08] mgvavb: †`Iqv Av‡Q, 3 a – ib = x – iy  a – ib = x3 – 3xy2 – i (3x2 y – y 3 )  a = x3 – 3xy2 ; b = 3x2 y – y 3  a + ib = x 3 – 3xy2 + i (3x2 y – y 3 ) = x 3 + 3x(iy)2 + 3x2 (iy) + (iy)3 = (x + iy)3  3 a + ib = x + iy