# Homework Help: Complex analysis showing solutions are inside or outside R

1. Apr 8, 2012

### d2j2003

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

Suppose w is not in the interval [-R,R] show that the equation z+$\frac{R^{2}}{z}$=2w has one solution z with |z|<R and one solution z with |z|>R

2. Relevant equations

none

3. The attempt at a solution

the book mentions that the quadratic is left unchanged by the substitution of $\frac{R^{2}}{z}$ for z which I understand.

It then goes on to say that because of this, the two roots z$_{1}$ and z$_{2}$are related by z$_{1}$z$_{2}$=R$^{2}$ which is where I get lost..

any help is appreciated. Thanks

2. Apr 8, 2012

### Dick

Write your equation as z^2-2wz+R^2=0. If there are two roots z1 and z2 then it factors into (z-z1)(z-z2)=0. Multiply that back out and equate it to the original equation.

3. Apr 8, 2012

### d2j2003

ahh ok.. can't believe I missed that. now the book says that because z1z2=R^2 the roots lie on opposite sides of the circle |z|=R. Why is this? i'm assuming they mean inside the circle and outside the circle when they say opposite sides.. And my guess would be that z1z2 = R^2 when z1=z2=R or if z1 is larger than R and z2 is smaller than R or vice versa. The latter meaning that z1 is inside the circle and z2 is outside. The first would mean that the roots lie on the circle|z|=R?

4. Apr 8, 2012

### Dick

|z1|*|z2|=R^2. So if |z1|<R, |z2|>R. That's the easy case. You are done. Now suppose |z1|=|z2|=R. To eliminate that possibility you'd better say something about w.

5. Apr 9, 2012

### d2j2003

so for |z1|=|z2|=R, w would have to be in [-R,R] but this possibility was eliminated in the statement of the problem. So then the only possibility is for |z1|<R and |z2|>R or vice versa.

6. Apr 9, 2012

### Dick

How did you show w would need to be in [-R,R]?

7. Apr 9, 2012

### d2j2003

because for |z1|=|z2|=R would mean that (z-w)$^{2}$=z$^{2}$-2zw+w$^{2}$=z$^{2}$-2zw+R$^{2}$ so w$^{2}$=R$^{2}$ and w=R which is in [-R,R]

8. Apr 9, 2012

### Dick

Why should (z-w)^2=z^2-2zw+R^2??? Your conclusion that w^2=R^2 is false. If you think harder about it you should be able to figure out the w is related to the sum of the roots z1 and z2, just like R^2 is related to the product of the roots.

9. Apr 9, 2012

### d2j2003

z1z2 must = R^2 and -z1-z2 must = -2w so z1+z2=2w now since |z1|=|z2|=R we have that R+R=2R=2w meaning R=w and R is in [-R,R]

10. Apr 9, 2012

### Dick

Yes, z1*z2=R^2 and z1+z2=2w. And yes, assume |z1|=|z2|=R. The rest of that doesn't make sense. w doesn't have to equal R.

11. Apr 9, 2012

### d2j2003

I thought if |z1|=|z2|=R then 2w=|z1|+|z2|=R+R=2R

12. Apr 9, 2012

### Dick

That doesn't work. Pick z1=iR and z2=(-iR). That makes w=0.

13. Apr 9, 2012

### d2j2003

ok.. so I know that z1*z2=R^2 and z1+z2=2w. And |z1|=|z2|=R I am just unsure of how to conclude that w is in [-R,R]...

thanks for your patience with this, its just not clicking.

14. Apr 9, 2012

### Dick

Can you convince me that z1 and z2 are complex conjugates of each other?

15. Apr 9, 2012

### d2j2003

could we say that since z1*z2=R^2 and R is a real number, that z1 and z2 have to be of the form (a+ib)(c-id) to eliminate the imaginary parts and since we are saying that |z1|=|z2|=R we know that a=c and b=d in the equation so z1 and z2 must be complex conjugates.

16. Apr 9, 2012

### Dick

That's not at all convincing. What must z1 and z2 look like in polar form?