What does ${(3 + i)}^{\frac{1}{3}}$ $(3+i)^(1/3)$ equal in a+bi form?

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What does ${(3 + i)}^{\frac{1}{3}}$ $(3+i)^(1/3)$ equal in a+bi form?

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Answer 1 · 2015-11-10T21:31:12

$\sqrt[6]{10} cos (\frac{1}{3} arctan (\frac{1}{3})) + \sqrt[6]{10} sin (\frac{1}{3} arctan (\frac{1}{3})) i$ $root(6)(10)cos(1/3 arctan(1/3)) + root(6)(10)sin(1/3 arctan(1/3)) i$

Explanation:

$3 + i = \sqrt{10} (cos (α) + i sin (α))$ $3+i = sqrt(10)(cos(alpha)+i sin(alpha))$ where $α = arctan (\frac{1}{3})$ $alpha = arctan(1/3)$

So

$\sqrt[3]{3 + i} = \sqrt[3]{\sqrt{10}} (cos (\frac{α}{3}) + i sin (\frac{α}{3}))$ $root(3)(3+i) = root(3)(sqrt(10))(cos(alpha/3)+i sin(alpha/3))$

$= \sqrt[6]{10} (cos (\frac{1}{3} arctan (\frac{1}{3})) + i sin (\frac{1}{3} arctan (\frac{1}{3})))$ $=root(6)(10)(cos(1/3 arctan(1/3)) + i sin(1/3 arctan(1/3)))$

$= \sqrt[6]{10} cos (\frac{1}{3} arctan (\frac{1}{3})) + \sqrt[6]{10} sin (\frac{1}{3} arctan (\frac{1}{3})) i$ $=root(6)(10)cos(1/3 arctan(1/3)) + root(6)(10)sin(1/3 arctan(1/3)) i$

Since $3 + i$ $3+i$ is in Q1, this principal cube root of $3 + i$ $3+i$ is also in Q1.

The two other cube roots of $3 + i$ $3+i$ are expressible using the primitive Complex cube root of unity $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2 i$ :

$ω (\sqrt[6]{10} cos (\frac{1}{3} arctan (\frac{1}{3})) + \sqrt[6]{10} sin (\frac{1}{3} arctan (\frac{1}{3})) i)$ $omega (root(6)(10)cos(1/3 arctan(1/3)) + root(6)(10)sin(1/3 arctan(1/3)) i)$

$= \sqrt[6]{10} cos (\frac{1}{3} arctan (\frac{1}{3}) + \frac{2 π}{3}) + \sqrt[6]{10} sin (\frac{1}{3} arctan (\frac{1}{3}) + \frac{2 π}{3}) i$ $=root(6)(10)cos(1/3 arctan(1/3) + (2pi)/3) + root(6)(10)sin(1/3 arctan(1/3)+(2pi)/3) i$

$ω^{2} (\sqrt[6]{10} cos (\frac{1}{3} arctan (\frac{1}{3})) + \sqrt[6]{10} sin (\frac{1}{3} arctan (\frac{1}{3})) i)$ $omega^2 (root(6)(10)cos(1/3 arctan(1/3)) + root(6)(10)sin(1/3 arctan(1/3)) i)$

$= \sqrt[6]{10} cos (\frac{1}{3} arctan (\frac{1}{3}) + \frac{4 π}{3}) + \sqrt[6]{10} sin (\frac{1}{3} arctan (\frac{1}{3}) + \frac{4 π}{3}) i$ $=root(6)(10)cos(1/3 arctan(1/3) + (4pi)/3) + root(6)(10)sin(1/3 arctan(1/3)+(4pi)/3) i$

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What does ${(3 + i)}^{\frac{1}{3}}$ $(3+i)^(1/3)$ equal in a+bi form?

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