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iren [92.7K]
3 years ago
6

All the cubes root of

} " alt=" \sqrt{3 + i} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
melisa1 [442]3 years ago
3 0

If you're looking for the cube roots of √(3 + <em>i </em>), you first have to decide what you mean by the square root √(…), since 3 + <em>i</em> is complex and therefore √(3 + <em>i </em>) is multi-valued. There are 2 choices, but I'll stick with 1 of them.

First write 3 + <em>i</em> in polar form:

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

Then the 2 possible square roots are

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

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

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

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

Then the 3 cube roots of √(3 + <em>i</em> ) are

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

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

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

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

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

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

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

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