Polynomials and complex numbers

[ivan_x3000]

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

 

[Mark44]

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?

 

[jbunniii]

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 $v$.

 

[Mark44]

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.

 

[ivan_x3000]

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

 

[Mark44]

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.

 

[ivan_x3000]

Sorry it is (u+iv)^-3 my bad

 

[jbunniii]

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.

 

[ivan_x3000]

Thank you so much that is a great hint haha

 

[krnysus]

@jbunniii can you please explain?
So (u + iv)^-1 = (u-iv)/9 then how do i solve it?

 

[Ray Vickson]

@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 $(u + iv)^{-1}$?

 

[krnysus]

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...

 

[jbunniii]

$$(u+iv)^{-3} = [(u+iv)^{-1}]^3 = \left(\frac{u-iv}{9}\right)^3 = ?$$