all the cubes root of \sqrt{3 + i} ​

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all the cubes root of
\sqrt{3 + i}
​

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RI SƠ 3 years 2021-08-24T04:56:12+00:00 1 Answers 3 views 0

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  1. diemthu
    0
    2021-08-24T04:57:25+00:00
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    If you’re looking for the cube roots of √(3 + i ), you first have to decide what you mean by the square root √(…), since 3 + i is complex and therefore √(3 + i ) is multi-valued. There are 2 choices, but I’ll stick with 1 of them.

    First write 3 + i in polar form:

    3 + i = √(3² + 1²) exp(i arctan(1/3)) = √10 exp(i arctan(1/3))

    Then the 2 possible square roots are

    • √(3 + i ) = ∜10 exp(i arctan(1/3)/2)

    • √(3 + i ) = ∜10 exp(i (arctan(1/3)/2 + π))

    and I’ll take the one with the smaller argument,

    √(3 + i ) = ∜10 exp(i arctan(1/3)/2)

    Then the 3 cube roots of √(3 + i ) are

    • ∛(√(3 + i )) = ¹²√10 exp(i arctan(1/3)/6)

    • ∛(√(3 + i )) = ¹²√10 exp(i (arctan(1/3)/6 + π/3))

    • ∛(√(3 + i )) = ¹²√10 exp(i (arctan(1/3)/6 + 2π/3))

    On the off-chance you meant to ask about the cube roots of 3 + i, and not √(3 + i ), then these would be

    • ∛(3 + i ) = ⁶√10 exp(i arctan(1/3)/3)

    • ∛(3 + i ) = ⁶√10 exp(i (arctan(1/3)/3 + 2π/3))

    • ∛(3 + i ) = ⁶√10 exp(i (arctan(1/3)/6 + 4π/3))

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