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Please help me with my Trigonometry?

Write each expression in the standard form for a complex number, a + bi.

1. (1 + i)^5

2. (-√3+i)5

4 Answers

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  • 2 years ago

    .

    ( 1 )

    (1 + i )⁵

    = ( √2 )⁵ [ cos(5 * ¼π ) + isin(5 * ¼π ) ]

    = 4√2 ( -½√2 - ½i√2 )

    = -4 - 4i

    = -4 (1 + i )

    ━━━━━

    ( 2 )

    Assuming (-√3 + i)⁵

    arctan (1/(-√3) ) = -⅙π = 7π/6

    (-√3 + i)⁵

    = ( 2 )⁵ [ cos(5 * 7π/6 ) + isin(5 * 7π/6 ) ]

    = 32 ( ½√3 - ½i )

    = 16( √3 - i )

    ━━━━━━

  • 2 years ago

    1.

    (1 + i)^5

    = -4 - 4 i

    2.

    (-√3 + i)^5

    = 27.7128129211020366964391414640939578710848840609660900488... +

    16 i

  • 2 years ago

    (1 +i)^5 =

    Use Pascal's triangle

    1^5 + 5*1^4*i + 10*1^3*i^2 + 10*1^2*i^3 + 5*1*i^4 + i^5 =

    1 + 5i -10 - 10i +5 + i =

    -4 - 4i

    (-√3 + i)^5 =

    (-√3)^5 + 5*(-√3)^4*i + 10*(-√3)^3*i^2 + 10*(-√3)^2*i^3 + 5*(-√3)*i^4 + i^5 =

    -9√3 + 45i +30√3 -30i -5√3 -i =

    16√3 +14i

  • ?
    Lv 7
    2 years ago

    (1 + i)^5 = (√2 e^(i*π/4))^5 = 4√2 e^(i*5π/4) = -4 - 4i

    or

    (1 + i)^5 = 1^5 + 5*1^4*i^1 + 10*1^3*i^2 + 10*1^2*i^3 + 5*1^1*i^4 + i^5 = -4 - 4i

    Use either of these methods again to answer question 2.

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