1. Feb 24, 2013

### nate9228

1. The problem statement, all variables and given/known data
Let A be complex, B be real. Show $\left|z^2\right|$+ Re(Az)+B=0 only has a solution if and only if $\left|A^2\right|$$\geq4B$. Then, assuming the above condition holds, show the solution is a circle or a single point.

2. Relevant equations

3. The attempt at a solution
Well, as above, I know the general quadratic equation for z, but I am not sure if the absolute value thing affects it and otherwise I am lost. My suspicion is that the solution will be a real number and since all of the coefficients are real, when using the quadratic equation, you need to have whats under the radical be positive. Something along those lines I think. I, however, have no clue how to prove it somewhat rigorously.

2. Feb 24, 2013

### Ferramentarius

$z = re^{i\phi}$ and $Re(z) = \frac{1}{2}(z+\bar{z})$.

3. Feb 24, 2013

### nate9228

4. Feb 24, 2013

### Dick

Put z=x+iy and write the equation out in terms of x and y. Try completing some squares. B must be real. It doesn't look like they want you to assume A is real though. Write it as Re(A)+i*Im(A).

Last edited: Feb 24, 2013
5. Feb 25, 2013

### nate9228

That is correct, A is complex, but only the real part of it is in the equation. One thing that is confusing me A LOT is that this is a quadratic, and I was under the impression quadratics ALWAYS have a solution which may be real or may be complex. For this problem I am to prove that the above equation has a solution if and only if the absolute value of (A^2) is greater than or equal to 4B. I dont even see how the absolute value of A comes into play. Oh and I am horrible at completing squares, I've never been formally taught how.

6. Feb 25, 2013

### Dick

Start by writing down what |z^2|+Re(Az)+B is in terms of x and y. And Re(Az) will contain Im(A).

7. Feb 25, 2013

### nate9228

Hey Dick, hopefully I'm not making you feel like your having to baby me. I am essentially teaching myself this material, my professor does not do much. Anyways, letting z= x+iy I know the equation becomes (x^2)+(y^2)+Re(A)x+B=0 and then I tried completing the square and get (y^2)+(x+ Re(A)/2)^2= ((Re(A)^2)/4)-B. And now I am not sure where to go.

8. Feb 25, 2013

### Dick

You are making great progress. Now since the left side is nonnegative then the right side must be too, right? But you need to fix up Re(Az). Az is (Re(A)+i*Im(A))*(x+iy). There's more to Re(Az) than just Re(A)x.

Last edited: Feb 25, 2013
9. Feb 25, 2013

### nate9228

Ahhhh, I naively assumed Re(Az)= Re(A)*Re(z)= Re(A)*x. The thought to multiply out Az first, then look at the real part of that product simply evaded me...I shall fix that now

10. Feb 25, 2013

### nate9228

Alrighty. So now I have, after correcting the Re(Az) error and a little simplifying to get the 2's out of the denominators as well as producing the required 4B, (4y-2Im(A))2+ (4x+ 2Re(A))2= (ReA)2+ (ImA)2- 4B = $\left|A^2\right|$-4B. Since the LHS is greater then or equal to 0 the RHS must be the same, hence the absolute value of A^2 must be greater then or equal to 4B. And now I just remembered this is an if and only if proof so now I assume I have to prove it the other way as well.

11. Feb 25, 2013

### Dick

Ok, so? Looks to me like that equation describes a circle in the xy plane. What are the center and radius?

12. Feb 25, 2013

### nate9228

Oh yeah, definitely a circle and if I remember the rules properly it is centered at (2Im(A), -2Re(A)) and r= √$\left|A^2\right|$-4B

13. Feb 25, 2013

### Dick

Close but not quite. And your previous answer wasn't quite right either. You are handling that factor of 4 wrong. A circle looks like (x-x0)^2+(y-y0)^2=r^2. Try it again more carefully.

14. Feb 25, 2013

### nate9228

Can you be more specific in regards to which part of my previous answer was wrong? Just so I know which part to look at. I see that I gave the ordered pair for the center backwards i.e (y,x) when it should be (x,y). I am not sure what you mean by handling the factor of 4 wrong though.

15. Feb 25, 2013

### nate9228

Ahh is it that I improperly multiplied the 4 through the equations? I think I should have multiplied through before I made the terms into squares, which would change for example 4y into 2y, etc.

16. Feb 25, 2013

### Dick

Yeah, the overall idea is fine. Just fix the numbers up a little.

17. Feb 25, 2013

### nate9228

Thank you

18. Feb 25, 2013

### Dick

Very welcome. Glad to see you figured out the 'completing the square' thing. You can do stuff with it you can't really pull off just using the quadratic equation when you are dealing with multiple variables.