Tiêu đề Complex Number Practice Sheet Solution HSC FRB 25.pdf ✅

RwUj msL ̈v  Final Revision Batch '25 1 Board Questions Analysis m„Rbkxj cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 1 1 1 1 1 1 1 1 1 2022 1 1 1 1 1 1 1 1 1 eûwbe©vPwb cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 4 4 4 4 5 4 4 4 5 2022 4 4 5 4 4 4 3 4 4 weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| DÏxcK-1: x = a + b + c 2 , y = a + b 2 + c DÏxcK-2: 7 + i8 = (p + iq)3 [XvKv †evW©- Õ23] (K) GK‡Ki GKwU KvíwbK Nbg~j  n‡j †`LvI †h,     1 +  + 3  6 = 64 (L) DÏxcK-1 Gi mvnv‡h ̈, hw` x 3 + y3 = 0 nq, Z‡e †`LvI †h, b = 1 2 (c + a) (M) DÏxcK-2 Gi mvnv‡h ̈ cÖgvY Ki †h, p 2 – q 2 = 7 4p + 2 q mgvavb: (K) L.H.S =     1 +  + 3  6 =      +  2 + 3  6 =     2  6 = 2 6  6 = 64 1 = 64 = R.H.S (Showed) (L) †`Iqv Av‡Q, x = a + b + c 2 , y = a + b 2 + c Ges x 3 + y3 = 0  x 3 = – y 3      x y 3 = – 1  x y = – 1, – , –  2  x = – y, – y, – y 2 GLb, x = – y a + b + c 2 = – a – b 2 – c  a + b + c 2 + a + b 2 + c = 0  2a + b ( +  2 ) + c ( +  2 ) = 0  2a = b + c  a = 1 2 (b + c) Avevi, x = – y  x + y = 0  a + b + c 2 + a + b 3 + c 2 = 0  2c 2 + a(1 + ) + b(1 + ) = 0  2c 2 – a 2 – b 2 = 0  2c = a + b  c = 1 2 (a + b) Avevi, x = – y 2  a + b + c 2 + a 2 + b 4 + c 3 = 0  2b + a (1 +  2 ) + c (1 +  2 ) = 0  2b – a – c = 0  2b = a + c  b = 1 2 (c + a) (Proved) (M) (p + iq)3 = 7 + i8 p 3 + 3p2 iq – 3pq2 – iq3 = 7 + i8 ev ̄Íe I KvíwbK Ask mgxK...Z K‡i, p 3 – 3pq2 = 7  p 2 – 3q2 = 7 p ..... (i) 3p2 q – q 3 = 8  3p2 – q 2 = 8 q ........ (ii) (i) + (ii)  p 2 – 3q2 + 3p2 – q 2 = 7 p + 8 q  4p2 – 4q2 = 7 p + 8 q  p 2 – q 2 = 7 4p + 2 q (Proved)

RwUj msL ̈v  Final Revision Batch '25 3 mgvavb: (K) (1 – i)–2 – (1 + i)–2 = 1 (1 – i) 2 – 1 (1 + i) 2 = 1 1 – 2i + i2 – 1 1 + 2i + i2 = 1 – 2i – 1 2i = – 1 – 1 2i = – 2 2i = – 1 i = i 2 i = i (Ans.) (L) †`Iqv Av‡Q,  = 45 ; r = 1 z = rcos + irsin = 1cos45 + isin45 = 1 2 + i 1 2 = 1 + i 2 z 2 = (1 + i) 2 2 = 1 + 2i + i2 2 = 2i 2 = i cÖ`Ë ivwk, z 8 + z6 + z4 + z2 + 1 = i4 + i3 + i2 + i + 1 = 0 + 1 [i + i2 + i3 + i4 = 0] = 1 (Ans.) (M) z = rcos + irsin  arg(z) = tan–1     rsin rcos = tan–1 tan = 0 z 2 = (rcos + irsin) 2 [cos2  – sin2  = cos2] = r2 cos2  + 2r2 icossin + i2 r 2 sin = r2 cos2  – r 2 sin2  + r2 isin2 = r2 cos2 + r2 isin2  arg(z2 ) = tan–1     r 2 sin2 r 2 cos2 = tan–1 tan2 = 2 = 2arg(z) (Proved) 5| Z1 = 1 – ix Ges Z2 = a + ib †hLv‡b a, b  [PÆMÖvg †evW©- Õ23] (K) x = 3 n‡j, Z1 †K †cvjvi AvKv‡i cÖKvk Ki| (L) cÖgvY Ki †h, x Gi GKwU ev ̄Íe gvb Z1 — Z1 – = Z2 – mgxKiY‡K wm× K‡i †hLv‡b a 2 + b2 = 1 (M) 3 Z2 = p + iq n‡j, cÖgvY Ki †h, – 2 (p2 + q2 ) = a p – b q mgvavb: (K) †`Iqv Av‡Q, x = 3  Z1 = 1 – i 3  |Z1| = 1 + ( 3) 2 = 2 Arg(Z1) = – tan–1    3 1 = –  3  Z1 Gi †cvjvi AvKvi 2       cos    –   3 + isin    –   3 = 2     cos  3 – isin  3 (L) Z1 = 1 – ix  Z1 – = 1 + ix Z2 = a + ib Z2 – = a – ib  Z1 — Z1 – = Z2 –  1 – ix 1 + ix = a – ib  1 + ix 1 – ix = 1 a – ib  1 + ix – 1 + ix 1 + ix + 1 – ix = 1 – a + ib 1 + a – ib [we‡qvRb-†hvRb]  2ix 2 = (1 – a + ib) (1 + a + ib) (1 + a) 2 – (ib) 2  ix = (1 + ib) 2 – a 2 1 + 2a + a2 + b2  ix = 1 + 2ib + i2 b 2 – a 2 1 + 2a + a2 + b2  ix = 1 + 2ib – (a 2 + b2 ) 1 + 2a + a2 + b2  ix = 1 + 2ib – 1 1 + 2a + 1 [†`Iqv Av‡Q a 2 + b2 = 1]  x = 2b 2 + 2a = b 1 + a hv GKwU ev ̄Íe msL ̈v (Proved) (M) Z2 = a + ib Ges 3 Z2 = p + iq  (a + ib) 1 3 = p + iq  a + ib = (p + iq)3  a + ib = p3 + 3p2 iq + 3pi2 q 2 + i3 q 3  a + ib = p3 + i3p2 q – 3pq2 – iq3 ev ̄Íe I KvíwbK Ask mgxK...Z K‡iÑ  a = p3 – 3pq2  b = 3p2 q – q 3 R.H.S = a p – b q = p 3 – 3pq2 p – 3p2 q – q 3 q = p2 – 3q2 – 3p2 – q 2 = – 2p2 – 2q2 = – 2 (p2 + q2 ) = L.H.S (Proved) 6| z = x + iy RwUj msL ̈vwUi AbyewÜ RwUj msL ̈v z –– | [wm‡jU †evW©- Õ23] (K) 4 – 49 Gi gvb wbY©q Ki| (L) x = 2 Ges y = 2 n‡j, z Gi eM©g~j wbY©q Ki| (M) |z + 4| – |z –– – 4| = 10 Øviv wb‡`©wkZ mÂvic‡_i mgxKiY wbY©q Ki|
4  Higher Math 2nd Paper Chapter-3 mgvavb: (K) awi, x = 4 – 49  x 4 = – 49 = (7i)2  (x2 ) 2 = (7i)2  x 2 =  7i = 7 2 ( 2i) = 7 2 (1  2 + i2 ) = 7 2 (1  i)2  x =  7 2 (1  i) (L) †`Iqv Av‡Q, z = x + iy ; x = 2 Ges y = 2  z = 2 + 2i awi, 2 + 2i = x + iy  2 + 2i = x2 + 2xiy + i2 y 2 ev ̄Íe I KvíwbK Ask mgxK...Z K‡i, x 2 – y 2 = 2 ....(i) Ges 2xy = 2  x 2 + y2 = (x 2 – y 2 ) 2 + 4x 2 y 2 = 2 2 + 22 = 2 2 ....(ii) (i) + (ii)  2x2 = 2 2 + 2  x =  2 + 1 (ii) – (i)  2y2 = 2 2 – 2  y =  2 – 1 2 + 2i =  ( 2 + 1 + i 2 – 1) (M) †`Iqv Av‡Q, z = x + iy  z –– = x – iy |z + 4| – |z –– – 4| = 10  |x + iy + 4| = 10 + |x – iy – 4|  (x + 4) 2 + y2 = 10 + (x – 4) 2 + y2  x 2 + 8x + 16 + y2 = 100 + 20 (x – 4) 2 + y2 + x 2 – 8x + 16 + y2  16x – 100 = 20 (x – 4) 2 + y2  4x – 25 = 5 (x – 4) 2 + y2  16x2 – 200x + 625 = 25(x2 – 8x + 16 + y2 )  9x2 + 25y2 = 225  x 2 5 2 + y 2 3 2 = 1 wb‡`©wkZ mÂvic_ hv GKwU Dce„‡Ëi mgxKiY| 7| z1 = – 1 – i 3, z2 = 3 – i. [ewikvj †evW©- Õ23] (K) z1 Gi eM©g~j wbY©q Ki| (L) †`LvI †h, Arg    z1 z2 = Arg z1 – Arg z2 (M) cÖgvY Ki †h,    1 2 z1 –– n +     1 2 z1 n = 2, hLb n Gi gvb 3 Øviv wefvR ̈ A_ev, – 1, hLb n Gi gvb Ab ̈ †Kv‡bv c~Y©msL ̈v| mgvavb: (K) z1 = – 1 – i 3 awi, z1 = – 1 – i 3 = x – iy  – 1 – i 3 = x 2 – 2ixy – y 2 ev ̄Íe I KvíwbK Ask mgxK...Z K‡i, x 2 – y 2 = – 1 ......(i) – 2xy = – 3 2xy = 3  x 2 + y2 = (x 2 – y 2 ) 2 + (2xy) 2 = 1 + 3 = 2......(ii) (i) + (ii)  2x2 = 1  x =  1 2 (ii) – (i)  2y2 = 3  y =  3 2 – 1 – i 3 =  1 2 (1 – 3i) (Ans.) (L) †`Iqv Av‡Q, z1 = – 1 – i 3 hv 3q PZzf©v‡M Aew ̄’Z  arg(z1) = –  + tan–1    – 3 – 1 = –  +  3 = – 2 3 z2 = 3 – i hv 4_© PZzf©v‡M Aew ̄’Z  arg(z2) = – tan–1     – 1 3 = –  6 z1 z2 = – 1 – i 3 3 – i = (– 1 – i 3) ( 3 + i) ( 3 – i) ( 3 + i) = – 3 – i – 3i – i 2 3 ( 3) 2 – i 2 = – 4i 4 = – i = 0 – i L.H.S = Arg    z1 z2 = – tan–1     – 1 0 = –  2 R.H.S = arg(z1) – arg(z2) = – 2 3 +  6 = – 4 +  6 = –  2  L.H.S = R.H.S (Showed) (M) †`Iqv Av‡Q, z1 = – 1 – i 3 1 2 z1 = – 1 – i 3 2 =  1 2 z1 –– = – 1 + i 3 2 =  2      1 2 z1 –– n +     1 2 z1 n = ( 2 ) n + () n