Title Complex Number Varsity Practice Sheet ‍Solution.pdf ✅

RwUj msL ̈v  Varsity Practice Sheet 3 e ̈vL ̈v:         1 + i 2 2 4n +         1 – i 2 2 4n =     1 – 1 + 2i 2 4n +     1 – 1 – 2i 2 4n = (i4 ) n + ((– i) ) 4 n = 1 + 1 = 2 22. i –2017 = ? –1 1 – i i DËi: – i e ̈vL ̈v: i –2017 = 1 i 2017 = 1 i = – i Note: i Gi Nv‡Zi †k‡li 2 wWwRU‡K 4 Øviv fvM K‡i cvIqv fvM‡klB n‡e i Gi NvZ| 23. i m + im+1 + im+2 + im+3 ? [m  Z] – 1 –i 0 i DËi: 0 e ̈vL ̈v: awi, m = 1  i 1 + i2 + i3 + i4 = 0 24. i 2 = – 1 n‡j i + i2 + i3 + ...... + i23 = ? [Agri Guccho 20-21] –1 i – i 1 DËi: –1 e ̈vL ̈v: i + i2 + i3 + ...... + i23 + i24 – i 24 = – i 24 = –1 Note: i Gi NvZ cici PviwU μwgK msL ̈v n‡j Zv‡`i †hvMdj 0 nq| 25. i i  i i 2  i i 3 ...... ii 37 = ? e  2 e –  2 1 i DËi: e –  2 e ̈vL ̈v: i + i2 + i3 + ...... + i36 + i37 = i37 = i GLb, i i  i i 2  i i 3 ...... ii 37 = i i + i2 + i3 + ...... + i37 = i i =     e i  2 i = e i 2  2 = e –  2     i = ei  2 Aqjv‡ii AvKvi 26. 1 + i + i2 + i3 + ..... i2014 + i2015 = ? 0 1 i –i DËi: 0 e ̈vL ̈v: 1 + i + i2 + i3 + i4 + i5 + i6 + i7 + .... + i2012 + i 2013 + i2014 + i2015 = (1 + i + i 2 + i3 ) + (1 + i + i2 + i3 ) + ..... + (1 + i + i 2 + i3 ) = 0 + 0 + ..... + 0 = 0 27. (1 + i) (1 + i2 ) (1 + i3 ) (1 + i4 ) = ? 0 2 5 6 DËi: 0 e ̈vL ̈v: 1 + i2 = 0  (1 + i) (1 + i2 ) (1 + i3 ) (1 + i4 ) = 0 28. z1 = 2 + i Ges z2 = 3 + i n‡j, z1z2 Gi gWzjvm = ? [JUST 14-15 ] 5 2 3 2 5 3 3 3 DËi: 5 2 e ̈vL ̈v: Mod(z1 z2) = Mod(z1)  Mod(z2) = 4 + 1  9 + 1 = 5  10 = 5 2 A_ev, z1 z2 = 6 + 2i + 3i + i2 = 5 + 5i | z1 z2| = 5 2 + 52 = 5 2 29. z1 = 1 – i I z2 = 3 + i n‡j z2 z1 Gi bwZÑ [DU 17-18] 5 12  6 –  4 – 5 12 DËi: 5 12 e ̈vL ̈v: Arg     z2 z1 = Arg(z2) – Arg(z1) = tan–     1 3 – (–tan–1 |1|) =  6 +  4 = 5 12 30. (a + ib) + (c + id) GB †hvMdj Gi gvb hw` ev ̄Íe msL ̈v nq, Zvn‡j †KvbwU mwVK? a + c = 0 b + d = 0 a + b = 0 b + c = 0 DËi: b + d = 0 e ̈vL ̈v: z1 = a + ib ; z2 = c + id z1 + z2 = (a + c) + i (b + d) GUv m¤ú~Y© ev ̄Íe n‡Z n‡j, KvíwbK Ask = 0  b + d = 0 31. If cos(logi4i) = a + ib, then a = 1, b = – 1 a = – 1, b = 1 a = 1, b = 0 a = 1, b = 2 DËi: a = 1, b = 0 e ̈vL ̈v: cos(4i log i) = a + ib  cos (4i log e ) i  2 = a + ib  cos     4i  i  2 log e = a + ib  cos (– 2) = a + ib  1 = a + ib  a = 1 ; b = 0
4  Higher Math 2nd Paper Chapter-3 32. 8 – i 1 + 8i = ? [CU 10-11] i – i 1 1 + i DËi: – i e ̈vL ̈v: 8 – i 1 + 8i = (8 – i)(1 – 8i) (1 + 8i)(1 – 8i) = 8 – 64i – i + 8i2 1 2 – (8i) 2 = – 65i 65 = – i 33. hw` 2 + 3i 2 – i = A + iB, †hLv‡b A I B ev ̄Íe msL ̈v, B = ? [Agri Guccho 19-20] 3 5 4 5 7 5 8 5 DËi: 8 5 e ̈vL ̈v: (2 + 3i)(2 + i) (2 – i)(2 + i) = 4 + 2i + 6i – 3 4 + 1 = 1 + 8i 5 = 1 5 + 8 5 i = A + Bi  B = 8 5 34. x Gi gvb †ei K‡iv hvi Rb ̈ (x – 2i) (1 + i) m¤ú~Y© KvíwbKÑ 2 – 2 4 – 4 DËi: – 2 e ̈vL ̈v: (x – 2i) (1 + i) = (x + 2) – i (x – 2) m¤ú~Y© Kvíwb‡Ki Rb ̈, x + 2 = 0  x = – 2 35. z = 1 cos – isin RwUj ivwk‡Z i Gi mnM KZ? sin – sin cos2  – sin2  – sincos DËi: sin e ̈vL ̈v: 1(cos + isin) (cos – isin)(cos + isin) = cos + isin cos2  + sin2  = cos + isin 36. 3 + 2isin hw` ev ̄Íe msL ̈v nq, Zvn‡j  = ? 2n n +  2 n None DËi: n e ̈vL ̈v: m¤ú~Y© ev ̄Íe n‡Z n‡j, KvíwbK Ask = 0 ev, sin = 0   = n 37. a + ib x + iy †K A + iB AvKv‡i cÖKvk Ki‡jÑ ax – by x 2 + y2 + i bx – ay x 2 + y2 ax + by x 2 + y2 + i bx – ay x 2 + y2 Both None DËi: ax + by x 2 + y2 + i bx – ay x 2 + y2 e ̈vL ̈v: (a + ib)(x – iy) (x + iy)(x – iy) = ax + by x 2 + y2 + i bx – ay x 2 + y2 38. x = a + ib , x2 = 3 + 4i , x3 = 2 + 11i n‡j (a + b) = ? 2 4 3 8 DËi: 3 e ̈vL ̈v: x = x 3 x 2 = (2 + 11i) (3 – 4i) (3 + 4i) (3 – 4i) = 6 – 8i + 33i + 44 25 = 2 + i  a = 2 ; b = 1 a + b = 3 39. [(cos + isin) (cos – isin)]–1 = ? i 1 – i – 1 DËi: 1 e ̈vL ̈v: [cos2  – i 2 sin2 ] –1 = (1)–1 = 1 40. (x + iy) (2 – 3i) = 4 + i n‡j x = ? I y = ? x = 8 13 , y = – 14 13 x = 5 13 , y = 14 13 x = 2 , y = 3 None DËi: x = 5 13 , y = 14 13 e ̈vL ̈v: (x + iy) (2 – 3i) = 4 + i  (2x + 3y) + i (– 3x + 2y) = 4 + i  2x + 3y = 4 ........ (i) – 3x + 2y = 1 ........... (ii) (i)  (3) + (ii)  (2)  9y + 4y = 12 + 2 = 14  y = 14 13 (i)  (2) – (ii)  (3)  4x + 9x = 8 – 3 = 5  x = 5 13 41. (1 + i)8 + (1 – i)8 = ? 16 – 16 32 – 32 DËi: 32 e ̈vL ̈v: (1 + i)8 = [(1 + i)2 ] 4 = (2i)4 = 16 (1 – i)8 = 16  (1 + i)8 + (1 – i)8 = 32 42. x = 1 + –1 2 n‡j, x 12 Gi gvb wb‡Pi †KvbwU? –1 1 – i i