How Do I Solve These Complex Number and Polynomial Problems?

[Niles]

Homework Statement

1) If I know that z_1 = (2+i) and z_2 = (2-i) are solutions to a polynomial, how do I find it? (I have six to chose between, it's a multiple choice). Do I just insert and see of it equals zero?

2) When I know that z^2 has modulus 4 and argument pi/2, how do I find modulus and argument for z?

3) If I know a polynomial (real numbers, not complex!) f(x) = 2x^2 and g(x) = cx, how do I find the number c, so g(x) is a tangent to the polynomial?

The Attempt at a Solution

1) I would just insert solutions.

2) I do not know how to approach this problem.

3) First, g(x) = f(x), and the gradient of g(x) must equal f'(x) in the point x_0?

 

[arildno]

1) Quite correct, although there might be additional clues in the actual polynomials given that makes the determination process easier.

2. You have:
z^{2}=4e^{i\pi(\frac{1}{2}+2n)}, n\in{Z}
Take the square root of this and see how many z's fits the expression.

3) I don't get your argument f(x)=g(x)?

 

[Niles]

1) Cool, thanks.

2) So modulus and argument for z is: mod = 2 and argument = pi/4?:

w_0 = 4^(1/2)*e^i*(pi/2)/2)

I don't have to find w_1, eh?

3) Nevermind, a silly question. I just had to read a section in my book.

 

[learningphysics]

There is another argument to consider for 2)...

 

[Niles]

pi/4 + 2kpi?

 

[learningphysics]

pi/4 + 2kpi?

Nope... that's the same angle. But you're close.

 

[Niles]

D'oh :-)

I hate to guess, but is it something like pi/4 +(2pi/n)*k = pi/4 + pi?

 

[learningphysics]

D'oh :-)

I hate to guess, but is it something like pi/4 +(2pi/n)*k = pi/4 + pi?

You got the right answer... but I'm kind of unsure about your left-handside formula... I'd do it like this using arildno's idea... you know that 2\theta = \pi/2 + 2n\pi then dividing by 2, \theta = \pi/4 + n\pi... you only need to consider n = 0 and n = 1... which give \pi/4 and 5\pi/4... all the other n's repeat the same angles.

 

[Niles]

Great, thanks!