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Answer:

[tex]{ \bf{b. \: \: {i}^{47} }}[/tex]

Step-by-step explanation:

[tex] {i}^{46} = ( {i}^{2} ) {}^{23} = - 1 \\ \\ {i}^{47} = (i)( {i}^{46} ) = ( - 1 \times i) = - i \\ \\ {i}^{48} = ( {i}^{2} ) {}^{24} = 1 \\ \\ {i}^{49} = (i)( {i}^{48}) = (1 \times i) = i[/tex]

Answer:

[tex]i^{46} = -i[/tex]

~~~~~~~~~~~~~~~~~~~

46/4 = 11 remainder 2 = [tex]-i[/tex]

47/4 = 11 remainder 3 = 1

48/4 = 11 remainder 0 =  [tex]i[/tex]

49/4 = 11 remainder 1 = -1

Step-by-step explanation:

[tex]i = \sqrt{-1} = i[/tex]  (remainder 0)

[tex]i^{2} = \sqrt{-1} * \sqrt{-1} = -1[/tex] (remainder 1)

[tex]i^{3} = \sqrt{-1} * \sqrt{-1} * \sqrt{-1}= -1 * \sqrt{-1} = -i[/tex] (remainder 2)

[tex]i^{4} = \sqrt{-1} * \sqrt{-1} * \sqrt{-1} * \sqrt{-1}= -1 * -1 = 1\\[/tex] (remainder 3)

[tex]i^{5} = \sqrt{-1} * \sqrt{-1} * \sqrt{-1} * \sqrt{-1}* \sqrt{-1}= 1 * \sqrt{-1} = i\\[/tex] (remainder 0)

The pattern repeats every FOUR in the exponent

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