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2  Higher Math 2nd Paper Chapter-3 11. KvíwbK msL ̈v i Ges n  N Gi Rb ̈ i 4n – i + i4n + 1 – 1 Gi gvb KZ? – i i 0 1 DËi: 0 e ̈vL ̈v: i 4n – i + i4n+1 – 1 = 1 – i + i4n  i – 1 [i4n = 1] = 0 – i + i = 0 12. i – 49 = ? – i i – 1 1 DËi: –i e ̈vL ̈v: i – 49 = 1 i 49 = 1 i = –i 2 i = – i 13. i –70 + 1 Gi gvb †KvbwU? 0 2 1 – i 1 + i DËi: 0 e ̈vL ̈v: 1 i 70 + 1 = 1 i 2 + 1 = 1 –1 + 1 = 0 14. i 2 = – 1 n‡j, – i – i –5 2i– 5 + i = ? –2 0 1 2 2 DËi: 0 e ̈vL ̈v: – i – i –5 2i– 5 + i = – i – 1 i 5 2 i 5 + i = – i – 1 i 2 i + i = –i 2 – 1 2 + i2 = + 1 – 1 2 – 1 = 0 15. –3 × –1 Gi gvb †KvbwU? 3i ± 3 – 3 3 DËi: – 3 e ̈vL ̈v: –3 × –1 = 3i.i = – 3 16. – 3 – 75 = ? 15 – 15 15i – 15i DËi: – 15 e ̈vL ̈v: 3  (–1)  75  (– 1) = 3i  75i = 225i 2 = –15 17.     1 – i 1 + i 2 = ? 1 – 1 2 1 2 –1 DËi: –1 e ̈vL ̈v:     1 – i 1 + i 2 = (1 – i) 2 (1 + i) 2 = 1 – 2i + i2 1 + 2i + i2 = –2i 2i = –1 18. n Gi me©wb¤œ †Kvb abvZ¥K c~Y©msL ̈vi Rb ̈ (1 + i)n = (1 – i)n nq? 4 8 2 16 DËi: 4 e ̈vL ̈v: (1 + i) n (1 – i) n = 1      1 + i 1 – i n = 1       (1 + i)(1 + i) 1 – i 2 n = 1      1 + 2i – i 2 1 + 1 n = 1  i n = 1 GLb, i 4 = i8 = i12 = ...... = 1  n Gi me©wb¤œ gvb 4 19. 2a = 1 + i n‡j a –12 = ? –1 1 i –i DËi: –1 e ̈vL ̈v: a = 1 + i 2  a 2 = 1 2 (1 + i)2 = 1 2  2i = i a –12 = 1 a 12 = 1 (a 2 ) 6 = 1 i 6 = 1 – = – 1 20. 2 n (1 – i) 2n + (1 + i) 2n 2 n = ? 2 0 2 (–i) n None of this DËi: 2 (–i) n e ̈vL ̈v: 2 n {(1 – i) 2 } n + {(1 + i) 2 } n 2 n = 2 n (–2i) n + (2i) n 2 n = 2 n –2 n  i n + 2 n i n 2 n = 1 (–i) n + in = 1 + in .(– i) n (– i) n = 1 + (– i 2 ) n (– i) n = 1 + 1 (– i) n = 2 (–i) n 21.     1 + i 2 8n +     1 – i 2 8n = ? 0 1 2 – 2 DËi: 2

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