How do you find the 3rd root of $8 e^{30 i}$ $8e^(30i)$ ?

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Answer 1 · 2016-09-16T23:14:39

$\sqrt[3]{8 e^{30 i}} = 2 e^{(10 + \frac{2 π}{3}) i}$ $root(3)(8e^(30i)) = 2e^((10+(2pi)/3)i)$

Note that $2 e^{10 i}$ $2e^(10i)$ is a third root of $8 e^{30 i}$ $8e^(30i)$ , but it may or may not be the principal one...

${(2 e^{10 i})}^{3} = 2^{3} \cdot e^{(10 i) \cdot 3} = 8 e^{30 i}$ $(2e^(10i))^3 = 2^3*e^((10i)*3) = 8e^(30i)$

To determine what the principal cube root is, we first need to determine which quadrant $8 e^{30 i}$ $8e^(30i)$ is in.

$\frac{30}{\frac{π}{2}} \approx 19.1 \equiv 3.1$ $30/(pi/2) ~~ 19.1 -= 3.1" "$ modulo $4$ $4$

This places $8 e^{30 i}$ $8e^(30i)$ in Q4, so its principal cube root is also in Q4.

$\frac{10}{\frac{π}{2}} \approx 6.4 \equiv 2.4$ $10/(pi/2) ~~ 6.4 -= 2.4" "$ modulo $4$ $4$

So $2 e^{10 i}$ $2e^(10i)$ is in Q3 and is not the principal cube root.

To get the principal cube root we can multiply by the primitive Complex cube root of $1$ $1$ :

$ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} = e^{\frac{2 π}{3} i}$ $omega = -1/2+sqrt(3)/2 = e^((2pi)/3i)$

This will result in a number in Q4

$2 e^{10 i} \cdot ω = 2 e^{10 i} e^{\frac{2 π}{3} i} = 2 e^{(10 + \frac{2 π}{3}) i}$ $2e^(10i) * omega = 2e^(10i) e^((2pi)/3i) = 2e^((10+(2pi)/3)i)$

How do you find the 3rd root of 8e30i8e^(30i)?