# Simple questions about complex numbers

1. Sep 1, 2007

### Niles

1. The problem statement, all variables and given/known data

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?

3. 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?

2. Sep 1, 2007

### 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)?

3. Sep 1, 2007

### 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.

4. Sep 1, 2007

### learningphysics

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

5. Sep 1, 2007

pi/4 + 2kpi?

6. Sep 1, 2007

### learningphysics

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

7. Sep 1, 2007

### Niles

D'oh :-)

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

8. Sep 1, 2007

### learningphysics

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.

9. Sep 1, 2007

### Niles

Great, thanks!