Polynomials and complex numbers

In summary: U + iv)^-3 = (u - iv)/9(u + iv)^2 ? I seriously have no clue how to express the imaginary part in terms of polynomial in v...$$(u+iv)^{-3} = [(u+iv)^{-1}]^3 = \left(\frac{u-iv}{9}\right)^3 = ?$$(U + iv)^-3 = (u - iv)/9
  • #1
ivan_x3000
19
0

Homework Statement


Suppose that u and v are real numbers for which u + iv has modulus 3. Express the imaginary part of (u + iv)^−3 in terms of a polynomial in v.

Homework Equations


The Attempt at a Solution


|u+iv|=3 then sort(u^2+i^2) = 3 then
u = 3 and v=0 or u=0 and v=3(0+3i)^-3

i swear i am missing something with these equations we've only being doing a topic on roots but wow it's different what I've been seeing in class
 
Last edited:
Physics news on Phys.org
  • #2
ivan_x3000 said:

Homework Statement


Suppose that u and v are real numbers for which u + iv has modulus 3. Express the imaginary part of (u + iv)−3 in terms of a polynomial in v.

Homework Equations


The Attempt at a Solution


|u+iv|=3 then sort(u^2+i^2) = 3 then
u = 3 and v=0 or u=0 and v=3(0+3i)^-3

i swear i am missing something with these equations we've only being doing a topic on roots but wow it's different what I've been seeing in class
If |u + iv| = 3, then u + iv is an arbitrary point on a circle of radius 3, centered at the origin in the complex plane. What effect does subtracting 3 from u + iv have on this circle?
 
  • #3
ivan_x3000 said:

Homework Statement


Suppose that u and v are real numbers for which u + iv has modulus 3. Express the imaginary part of (u + iv)−3 in terms of a polynomial in v.


Homework Equations





The Attempt at a Solution


|u+iv|=3 then sort(u^2+i^2) = 3 then
u = 3 and v=0 or u=0 and v=3
Those are two solutions, but there are infinitely many more.

By the way, is there a typo? The imaginary part of ##(u+iv) - 3## is simply ##v##, which is already a polynomial in ##v##.
 
  • #4
jbunniii said:
Those are two solutions, but there are infinitely many more.

By the way, is there a typo? The imaginary part of ##(u+iv) - 3## is simply ##v##, which is already a polynomial in ##v##.
The way I'm interpreting the problem, u + iv - 3 doesn't represent just a single point.
 
  • #5
Mark44 said:
The way I'm interpreting the problem, u + iv - 3 doesn't represent just a single point.

Yes definitely a type it was meant to be (u+iv)^3
 
  • #6
ivan_x3000 said:
Yes definitely a type it was meant to be (u+iv)^3
In the OP you changed it to (u + iv)-3 and above you have (u + iv)3. Which is it?

Really, you need to be more careful. If we don't know what the problem is, we can't help you.
 
  • #7
Sorry it is (u+iv)^-3 my bad
 
  • #8
OK, as hint I would recommend starting with
$$(u+iv)(u-iv) = |u+iv|^2$$
Rearrange this to get a formula for ##(u+iv)^{-1}## and use what you know about the modulus.
 
  • Like
Likes 1 person
  • #9
Thank you so much that is a great hint haha
 
  • #10
@jbunniii can you please explain?
So (u + iv)^-1 = (u-iv)/9 then how do i solve it?
 
  • #11
krnysus said:
@jbunniii can you please explain?
So (u + iv)^-1 = (u-iv)/9 then how do i solve it?

Can you really not see how to get ##(u + i v)^{-3}## if you know ##(u + iv)^{-1}##?
 
  • #12
Ray Vickson said:
Can you really not see how to get ##(u + i v)^{-3}## if you know ##(u + iv)^{-1}##?

(U + iv)^-3 = (u - iv)/9(u + iv)^2 ? I seriously have no clue how to express the imaginary part in terms of polynomial in v...
 
  • #13
$$(u+iv)^{-3} = [(u+iv)^{-1}]^3 = \left(\frac{u-iv}{9}\right)^3 = ?$$
 

Related to Polynomials and complex numbers

1. What are polynomials?

Polynomials are algebraic expressions that consist of variables and coefficients, combined using the operations of addition, subtraction, and multiplication. They can also include exponents, but the variables must have whole number exponents.

2. What is the degree of a polynomial?

The degree of a polynomial is the highest exponent of the variable in the expression. For example, the degree of the polynomial 5x3 + 2x2 + 7x + 1 is 3, since that is the highest exponent of x.

3. How do you add and subtract polynomials?

To add or subtract polynomials, you must combine like terms. Like terms are terms that have the same variable with the same exponent. You can add or subtract the coefficients of these like terms, while keeping the variable and exponent the same.

4. What are complex numbers?

Complex numbers are numbers that are expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, which is equal to the square root of -1. These numbers allow us to work with and solve certain equations that cannot be solved using only real numbers.

5. How do you multiply and divide complex numbers?

To multiply complex numbers, you can use the FOIL method, just as you would for binomials. To divide complex numbers, you must first rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator. Then, you can simplify the resulting expression.

Similar threads

  • Calculus and Beyond Homework Help
Replies
27
Views
772
  • Calculus and Beyond Homework Help
Replies
5
Views
561
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
836
  • Calculus and Beyond Homework Help
Replies
18
Views
2K
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
328
Replies
1
Views
629
  • Calculus and Beyond Homework Help
Replies
9
Views
5K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
Back
Top