Tiêu đề Complex Number Engg Practice Sheet Solution.pdf ✅

RwUj msL ̈v  Engineering Practice Sheet Solution 1 03 RwUj msL ̈v Complex Number WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1| z1 = – i + 3 I z2 = 1 + 3i n‡j z1 I – z2 Gi ga ̈eZ©x †KvY KZ? [BUET 23-24] mgvavb: z1 = 3 – i – z2 = 1 – 3i arg(z1) = – tan–1     1 3 = –  6 arg( – z2) = – tan–1 ( 3) = –  3  arg(z1) – arg( – z2) = –  6 +  3 =  6 2| z, iz, z + iz wÎfz‡Ri kxl© we›`y n‡j, wÎfz‡Ri †ÿÎdj KZ? [BUET 22-23] mgvavb: X Y z iz z +iz    =  1 2 | z | | z | = 1 2 (| z |) 2 eM©GKK 3| –1 + 3i †K †cvjvi ̄’vbv1⁄4 iƒ‡c cÖKvk Ki| G‡`i gWzjvm I Av ̧©‡g›U wbY©q Ki| [BUET 20-21] mgvavb: gWzjvm, r = (–1) 2 + ( 3) 2 = 2 Av ̧©‡g›U,  =  – tan–1    3 1 =  –  3 = 2 3  †cvjvi ̄’vbv1⁄4     2 2 3 (Ans.) 4| z + i z + 2 we›`yi mÂvic‡_i mgxKiY wbY©q Ki, hLb GwU m¤ú~Y© KvíwbK| [BUET 19-20; MIST 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| 5| hw` rei = 3 + 2i 2 + 3i + 1 + 5i 1 – 2i nq, Z‡e r I  Gi gvb wbY©q Ki| [BUET 17-18] mgvavb: rei = 3 + 2i 2 + 3i + 1 + 5i 1 – 2i  rei = –57 65 + 66 65 i  r =     –57 65 2 +     66 65 2 = 3 5   =  – tan–1       66 65 57 65 =  – tan–1     66 57 (Ans.) 6| 1 + 2i 1 – 3i †K r (cos + i sin) AvKv‡i cÖKvk Ki| [BUET 16-17] mgvavb: 1 + 2i 1 – 3i = (1 + 2i) (1 + 3i) (1 – 3i) (1 + 3i) = – 1 2 + 1 2 i  r =     – 1 2 2 +     1 2 2 = 1 2  = tan–1 1 2 – 1 2 = tan–1 (–1) = 135  1 + 2i 1 – 3i = 1 2 (cos + i sin) †hLv‡b,  = 135
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