Solving cubic roots (with imaginary number)

In summary, the conversation discusses solving for complex numbers in polar form and converting imaginary numbers to polar form. The participants mention using the polar form of an imaginary number and converting to polar coordinates. However, they are unsure of how to apply this to the given equation and are seeking assistance.
  • #1
jeff1evesque
312
0

Homework Statement


(0.1 - 0.3j)^(1/3) = a + bj, where j is the imaginary number or more specifically sqrt(-1).
Does anyone know how to solve for a and b?


Homework Equations


I've looked at cubic function equations, along with some polar equations. However, the latter requires some angle to be involved, and I have no such angle. When I try to solve the problem straight out, I get something really crazy.


The Attempt at a Solution


For instance, I get:
(a + bj)^3 = a^3 + 3a^2bi + 3ab^2 - b^3i = 0.1 - 0.3j

thanks,


JL
 
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  • #2
You aren't supposed to treat it as a cubic function, that would be crazy.

They're all complex numbers, so what happens when you change them all to polar form?
 
  • #3
queenofbabes said:
You aren't supposed to treat it as a cubic function, that would be crazy.

They're all complex numbers, so what happens when you change them all to polar form?

That's the part I would like to know I guess- which I am not sure about how to do. Could you help me, doing it the other way as I expanded the polynomial was just impossible.
 
  • #4
Do you know how to convert to polar form?

What is 0.1 - 0.3j in polar form?
Therefore what is (0.1-0.3j)^1/3 in polar form?
 
  • #5
queenofbabes said:
Do you know how to convert to polar form?

What is 0.1 - 0.3j in polar form?
Therefore what is (0.1-0.3j)^1/3 in polar form?

to convert to polar form we convert the following:
x = rcos(theta),
y = rsin(theta),
[tex]z = sqrt(x^2 + y^2)[/tex]

But I can't see how this would convert our equation to polar coordinates. So I guess I don't know how to do the conversion.
 
  • #6
I guess not. I'm referring to the polar form of an imaginary number, expressed as z = r e^(theta)

so for z = a + ib
r = sqrt (a^2 + b^2)
tan (theta) = b/a

It should be in your textbook somewhere...
 
  • #7
queenofbabes said:
I guess not. I'm referring to the polar form of an imaginary number, expressed as z = r e^(theta)

so for z = a + ib
r = sqrt (a^2 + b^2)
tan (theta) = b/a

It should be in your textbook somewhere...

Oh yea, I saw this on wikipedia earlier today, but didn't know how to use it. I've never seen such an equation till today. My multivariate calculus course I don't think covered this.
 
  • #8
This isn't calculus. This is complex numbers...
 

What is the process for solving cubic roots with imaginary numbers?

The process for solving cubic roots with imaginary numbers involves using the cubic formula, which is derived from the quadratic formula. The cubic formula is:
x = (-b ± √(b^2 - 4ac)) / 2a
where a, b, and c are the coefficients of the cubic equation. This formula will result in three possible solutions, two of which will be complex (involving imaginary numbers).

What are imaginary numbers and how are they used in solving cubic roots?

Imaginary numbers are numbers that cannot be represented on the number line and are denoted by the letter i. They are used in solving cubic roots because in some cases, the solutions to cubic equations will involve taking the square root of a negative number, which is where imaginary numbers come into play. They allow us to represent and work with these complex solutions.

Can cubic roots have multiple solutions?

Yes, cubic roots can have up to three solutions. This is because a cubic equation is a polynomial of degree 3, which means it can have at most 3 roots. However, some equations may have repeated roots, resulting in fewer than 3 distinct solutions.

How can I check my answer when solving cubic roots with imaginary numbers?

To check your answer when solving cubic roots with imaginary numbers, you can plug the solutions back into the original equation and see if they satisfy the equation. You can also use a graphing calculator to graph the equation and see if the solutions correspond to the x-intercepts.

Are there any tips for solving cubic roots with imaginary numbers?

One tip for solving cubic roots with imaginary numbers is to first simplify the equation as much as possible before using the cubic formula. This can help make the calculations easier and reduce the risk of errors. It is also important to remember to include the ± symbol when finding the solutions, as there will be two possible solutions for the complex roots.

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