MHB Complex numbers and conjugates

[Sean1]

Hi everyone,

Can you please assist with the following problem?

The complex numbers z and w are such that for the real variable x,
(x-z)(x-w)=ax²+bx+c for real a,b and c.

By letting z=p+qi and w=r+si, prove that z and w must be conjugates of one another.So far, I have determined that a=1, -b=w+z and c=wz,
I know I need to show that q+s=0 and that p=r, but I am not sure how to proceed.

Thanks for reading my post.

 

[Sudharaka]

Hi everyone,

Can you please assist with the following problem?

The complex numbers z and w are such that for the real variable x,
(x-z)(x-w)=ax²+bx+c for real a,b and c.

By letting z=p+qi and w=r+si, prove that z and w must be conjugates of one another.So far, I have determined that a=1, -b=w+z and c=wz,
I know I need to show that q+s=0 and that p=r, but I am not sure how to proceed.

Thanks for reading my post.

Hi Sean,

The three equations you have obtained are correct. Substitute for $w$ and $z$ in these equations. For example,

\[-b=w+z\Rightarrow (p+qi)+(r+si)=-b\Rightarrow (p+r)+i(q+s)=-b\]

Since $-b$ is a real number what can you say about $q+s$ which is the imaginary part?

 

[Prove It]

Hi everyone,

Can you please assist with the following problem?

The complex numbers z and w are such that for the real variable x,
(x-z)(x-w)=ax²+bx+c for real a,b and c.

By letting z=p+qi and w=r+si, prove that z and w must be conjugates of one another.So far, I have determined that a=1, -b=w+z and c=wz,
I know I need to show that q+s=0 and that p=r, but I am not sure how to proceed.

Thanks for reading my post.

Remember that b and c are real. The only way you can add two complex numbers to get a real number is if the imaginary parts cancel out, so $\displaystyle \begin{align*} \mathcal{I}\,(z) = -\mathcal{I}\,(w) \end{align*}$.

Now if you write $\displaystyle \begin{align*} z = p + \mathrm{i}\,q \end{align*}$ and $\displaystyle \begin{align*} w = r + \mathrm{i}\,s \end{align*}$ (where p,q,r,s are all real), then we have already shown that s = -q, giving $\displaystyle \begin{align*} w = r - \mathrm{i}\,q \end{align*}$. Multiplying z and w gives

$\displaystyle \begin{align*} w\,z &= \left( r - \mathrm{i}\,q \right) \left( p + \mathrm{i}\,q \right) \\ &= p\,r + \mathrm{i}\,q\,r - \mathrm{i}\,p\,q - \mathrm{i}^2\, q^2 \\ &= p\,r + q^2 + \mathrm{i} \, \left( q\,r - p\,q \right) \end{align*}$

For this to be real,

$\displaystyle \begin{align*} q\,r - p\,q &= 0 \\ q \, \left( r - p \right) &= 0 \\ q = 0 \textrm{ or } r - p &= 0 \\ q = 0 \textrm{ or } p &= r \end{align*}$

Therefore, if w and z are nonreal, $\displaystyle \begin{align*} \mathcal{R}\,(z) = \mathcal{R}\,(w) \end{align*}$.

This has shown that w and z must be complex conjugates.