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MATEMATICA BASICA II NUMEROS COMPLEJOS EJERCICIO 12.(− 128 +128 √ 3 i ) 1 8 r= √ ( − 128 ) 2 + ( 128 √ 3 ) 2 = 256 θ = tan −1 128 √ 3 − 128 = 60 (− 128 +128 √ 3 i ) 1 8 = 256 1 8 [ cos 1 8 ( 60 +2 πk ) +isen 1 8 ( 60 +2 πk ) ] Para: 60 +2 π ( 0 ) 60 +2 π ( 0 ) +isen 1 8 ¿ cos 1 8 ¿ k = 0 → z 1 = 2 ¿ k = 0 →z 1 = 2 [ cos 1 8 ( 60 ) +isen 1 8 ( 60 ) ] k = 1 → z 2 = 2 [ cos 1 8 ( 60 +2 π ( 1 ) )+ isen 1 8 ( 60 +2 π ( 1 ) ) ] k = 1 → z 2 = 2 [ cos 1 8 ( 52.5 )+ isen 1 8 ( 52.5 ) ] k = 2 → z 3 = 2 [ cos 1 8 ( 60 +2 π ( 2 ) )+ isen 1 8 ( 60 +2 π ( 2 ) ) ] k = 2 → z 3 = 2 [ cos 1 8 ( 97.5 )+ isen 1 8 ( 97.5 ) ] k = 3 → z 4 = 2 [ cos 1 8 ( 60 +2 π ( 3 ) )+ isen 1 8 ( 60 +2 π ( 3 ) ) ]

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