( 1 + i ) 2 = 1 + i 2 + 2 i = 1 − 1 + 2 i = 2 i {(1 + i)^2} = 1 + {i^2} + 2i = 1 - 1 + 2i = 2i\\ ( 1 + i ) 2 = 1 + i 2 + 2 i = 1 − 1 + 2 i = 2 i ( 1 − i ) 2 = 1 + i 2 − 2 i = 1 − 1 − 2 i = − 2 i {(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i\\ ( 1 − i ) 2 = 1 + i 2 − 2 i = 1 − 1 − 2 i = − 2 i
− 1 2 + i 3 2 = ω - {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega − 2 1 + 2 i 3 = ω ⇒ − 1 + i 3 = 2 ω \Rightarrow - 1 + i\sqrt 3 = 2\omega ⇒ − 1 + i 3 = 2 ω
and − 1 2 − i 3 2 = ω 2 - {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2} − 2 1 − 2 i 3 = ω 2 ⇒ − 1 − i 3 = 2 ω 2 \Rightarrow - 1 - i\sqrt 3 = 2{\omega ^2} ⇒ − 1 − i 3 = 2 ω 2 ⇒ 1 + i 3 = − 2 ω 2 \Rightarrow 1 + i\sqrt 3 = - 2{\omega ^2}\\ ⇒ 1 + i 3 = − 2 ω 2 K = ( − 1 + i 3 ) 21 ( 1 − i ) 24 + ( 1 + i 3 ) 21 ( 1 + i ) 24 K = {{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 - i} \right)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 + i} \right)}^{24}}}}\\ K = ( 1 − i ) 24 ( − 1 + i 3 ) 21 + ( 1 + i ) 24 ( 1 + i 3 ) 21 = ( 2 ω ) 21 ( ( 1 − i ) 2 ) 12 + ( − 2 ω ) 21 ( ( 1 + i ) 2 ) 12 = {{{{(2\omega )}^{21}}} \over {{{\left( {{{(1 - i)}^2}} \right)}^{12}}}} + {{{{( - 2\omega )}^{21}}} \over {{{\left( {{{(1 + i)}^2}} \right)}^{12}}}}\\ = ( ( 1 − i ) 2 ) 12 ( 2 ω ) 21 + ( ( 1 + i ) 2 ) 12 ( − 2 ω ) 21 = 2 21 . ω 21 ( − 2 i ) 12 + ( − 2 ) 21 ( ω 2 ) 21 ( 2 i ) 12 = {{{2^{21}}.{\omega ^{21}}} \over {{{( - 2i)}^{12}}}} + {{{{( - 2)}^{21}}{{({\omega ^2})}^{21}}} \over {{{(2i)}^{12}}}} = ( − 2 i ) 12 2 21 . ω 21 + ( 2 i ) 12 ( − 2 ) 21 ( ω 2 ) 21 ω 3 = 1 {\omega ^3} = 1 ω 3 = 1 i 4 = 1 {i^4} = 1 i 4 = 1
= 2 21 2 12 − 2 21 2 12 = 0 \\ = {{{2^{21}}} \over {{2^{12}}}} - {{{2^{21}}} \over {{2^{12}}}} = 0 = 2 12 2 21 − 2 12 2 21 = 0
∴ n = [ ∣ K ∣ ] = [ ∣ 0 ∣ ] = 0 \\ \therefore n = \left[ {|K|} \right] = \left[ {|0|} \right] = 0\\ ∴ n = [ ∣ K ∣ ] = [ ∣0∣ ] = 0 ∑ j = 0 5 ( j + 5 ) 2 − ∑ j = 0 5 ( j + 5 ) \sum\limits_{j = 0}^5 {{{(j + 5)}^2}} - \sum\limits_{j = 0}^5 {(j + 5)} \\ j = 0 ∑ 5 ( j + 5 ) 2 − j = 0 ∑ 5 ( j + 5 ) = ∑ j = 0 5 ( j 2 + 25 + 10 j − j − 5 ) = \sum\limits_{j = 0}^5 {({j^2} + 25 + 10j - j - 5)} \\ = j = 0 ∑ 5 ( j 2 + 25 + 10 j − j − 5 ) = ∑ j = 0 5 ( j 2 + 9 j + 20 ) = \sum\limits_{j = 0}^5 {({j^2} + 9j + 20)}\\ = j = 0 ∑ 5 ( j 2 + 9 j + 20 ) = ∑ j = 0 5 j 2 + 9 ∑ j = 0 5 j + 20 ∑ j = 0 5 1 = \sum\limits_{j = 0}^5 {{j^2}} + 9\sum\limits_{j = 0}^5 {j + 20\sum\limits_{j = 0}^5 1 }\\ = j = 0 ∑ 5 j 2 + 9 j = 0 ∑ 5 j + 20 j = 0 ∑ 5 1 = 5 × 6 × 11 6 + 9 ( 5 × 6 2 ) + 20 × 6 = {{5 \times 6 \times 11} \over 6} + 9\left( {{{5 \times 6} \over 2}} \right) + 20 \times 6\\ = 6 5 × 6 × 11 + 9 ( 2 5 × 6 ) + 20 × 6 = 55 + 135 + 120 = 55 + 135 + 120\\ = 55 + 135 + 120 = 310 = 310 = 310
