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

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In summary, the conversation discusses a quadratic equation and the options for solving it. The equation is z^2+(p+iq)z+r+is=0, where p,q,r,s are real and non-zero root. The options are 1. pqr=r^2+p^2s, 2. prs=q^2+r^2p, 3. qrs=p^2+s^2q, and 4. pqs=s^2+q^2r. The conversation also mentions using the quadratic formula to solve for z.
  • #1
juantheron
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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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  • #2


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


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.
 
  • #4


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 ...
 

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

1. What does the equation represent?

The equation represents a quadratic equation in the complex number system. It can be used to solve for the values of z that make the equation true.

2. What do the variables in the equation stand for?

The variable z represents a complex number, while p, q, r, and s represent real numbers. The values of p, q, r, and s determine the coefficients of the equation.

3. How do you solve this equation?

To solve this equation, you can use the quadratic formula for complex numbers:
[tex]z = \frac{-(p+iq) \pm \sqrt{(p+iq)^2 - 4(r+is)}}{2}[/tex]
You can also use factoring or completing the square to solve for z.

4. What are the roots of this equation?

The roots of this equation are the values of z that make the equation true. There can be either two distinct roots, one double root, or no real roots, depending on the values of the coefficients.

5. How is this equation used in real life?

This type of equation is often used in engineering, physics, and other scientific fields to model and solve problems involving complex numbers. It can also be used in economics and finance to model and analyze complex systems.

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