Finding Principal Root of \sqrt[3]{8i}: Stuck at Arctan?

In summary, the conversation discusses finding the principal root of \sqrt[3]{8 i} and converting it to polar coordinates. The point (0, 8) is located on the positive imaginary axis, making \theta = \frac{\pi}{2} or 90 degrees. The formula \theta = \arctan \frac{y}{x} can also be used, but it may be undefined in this case.
  • #1
MikeH
29
0
I have to find the principal root of [tex]\sqrt[3]{8 i}[/tex]
But I get stuck at this part
change this to polar coordinates...
[tex]r= \sqrt {x^2 + y^2} [/tex]
which makes [tex]r=8[/tex]
but when I try to find [tex]\theta[/tex]
[tex]\theta = \arctan \frac{y}{x}[/tex]
from the original x = 0 so how do I find [tex]\theta[/tex]?
 
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  • #2
Don't just use formulas without thinking! When you say change "this" to polar coordinates you mean 8i: in the complex plane, that's the point (0, 8)- on the positive imaginary (y) axis which makes a right angle with the real (x) axis- [itex]\theta[/itex] is [itex]\frac{\pi}{2}[/itex] or 90 degrees.

(Of course, [itex]\theta= arctan\frac{y}{x}[/itex] does work even in this case: [itex]tan(\frac{\pi}{2}) [/itex] is undefined.}
 
  • #3
Thanks for your help, I found the answer :smile:
 

1. What is the principal root of 38i?

The principal root of 38i is the number that, when cubed, results in 8i. In this case, the principal root is 2i.

2. How do I find the principal root of 38i?

To find the principal root of 38i, you can use the formula 3z = r(cos(θ + 2πn) + i sin(θ + 2πn)), where z is the number inside the cube root, r is the cube root of the magnitude of z, θ is the argument of z, and n is any integer. In this case, z = 8i, so r = 2, θ = π/2, and n = 0. Plugging these values into the formula gives us the principal root of 2i.

3. What is Arctan and how does it relate to finding the principal root of 38i?

Arctan is the inverse function of tangent, which is used to find the angle between the x-axis and a line passing through the origin and a point on the unit circle. In order to find the principal root of 38i, we can use the arctan function to find the argument of z (in this case, θ) and then plug it into the formula for finding the principal root.

4. Why am I stuck at Arctan when trying to find the principal root of 38i?

It is common to get stuck at the arctan step when trying to find the principal root of 38i because the argument of z (in this case, θ) can have multiple values. This is because tangent has a period of π and can take on multiple values for a given input. It is important to remember to add 2πn (where n is any integer) to the argument found using arctan in order to account for all possible values and find the correct principal root.

5. Is there a shortcut or easier way to find the principal root of 38i?

Yes, there is a shortcut that can be used to find the principal root of 38i. Instead of using arctan, you can use the formula 3z = 3r(cos(θ/3) + i sin(θ/3)), where z is the number inside the cube root, r is the cube root of the magnitude of z, and θ is the argument of z. This formula allows you to find the principal root without having to use arctan to find θ.

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