# Given that the equation $$z^2+(p+iq)z+r+is=0,$$ where

1. Jul 9, 2011

### juantheron

Given that the equation $$z^2+(p+iq)z+r+is=0,$$ where $$p,q,r,s$$ are real and non-zero root then
which one is right
1. $$pqr=r^2+p^2s$$

2. $$prs=q^2+r^2p$$

3. $$qrs=p^2+s^2q$$

4. $$pqs=s^2+q^2r$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 9, 2011

### SammyS

Staff Emeritus
Re: Complexnumber

What have you tried?

Where are you stuck?

To start, let z = x + y i , where x & y are real.

I assume i 2 = -1

3. Jul 10, 2011

### juantheron

Re: Complexnumber

put $$z=x+iy$$$$(x+iy)^2+(p+iq)(x+iy)+r+is=0$$ $$x^2-y^2+2ixy+px-qy+i(qx+py)+r+is=0$$$$\left(x^2-y^2+px-qy+r\right)+i\left(qx+py+2xy+s\right)=0+i.0$$$$\left(x^2-y^2+px-qy+r\right)=0$$$$\left(qx+py+2xy+s\right) = 0$$

Now I am struck here.

4. Jul 10, 2011

### SammyS

Staff Emeritus
Re: Complexnumber

Sorry for my lame suggestion !

Use the quadratic formula to solve for z, then the part of the instructions which seem to be missing some words; in bold below.

... $p,q,r,s$ are real and non-zero root ...