Given that the equation [tex]z^2+(p+iq)z+r+is=0,[/tex] where

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

2. [tex]prs=q^2+r^2p[/tex]

3. [tex]qrs=p^2+s^2q[/tex]

4. [tex]pqs=s^2+q^2r[/tex]
 
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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
 


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

Now I am struck here.
 


juantheron said:
Given that the equation [tex]z^2+(p+iq)z+r+is=0,[/tex] where [tex]p,q,r,s[/tex] are real and non-zero root then
which one is right
...
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.

... [itex]p,q,r,s[/itex] are real and non-zero root ...