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

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.)