Complex conjugate and their product

Question

I have two complex numbers that are non real, $k$ and $z$. $k$ and $z$ are going to be complex conjugates if and only if the product $(x-k)(x-z)$ is a polynomial with real coefficients.

Here is my answer :

$$k=a+bi\\ z=c+di\\ (x-k)(x-z) = x^2-(k+z)x+kz$$

After that, I'm not sure what to do. I guess that I need to add $k+z$ and multiply $k*z$.

Any hints?

Answer 1

So, $k+z=(a+c)+(b+d)i$ and $kz = (ac-bd)+(ad+bc)i$. Both being real is equivalent to $b+d=0$ and $ad+bc=0$, which in turn is equivalent to $b=-d$ and $a=c$, which is equivalent to $k$ and $z$ being complex conjugate.

Answer 2

The requirement is that $a,b,c,d$ are real numbers such that $k,z$ are representing complex numbers. The sum of real numbers is also a real number and a product of real numbers is also a real number (see also: "field axioms").

Take the product (x-k)(x-z) and simplify then you will get the answer.

Answer 3

Okay, look at it this way.

Let $ax^2 + bx + c$ be a 2nd degree polynomial with real coefficients and no real roots. But it does have complex roots. What are the roots?

Well by the quadratic equation the are $z = \frac {-b}{2a} + \frac {\sqrt{4ac - b^2}}{2a} = \frac {-b}{2a} + \frac {\sqrt{b^2 - 4ac}}{2a}i$ (as there are no real roots, we know $b^2 > 4ac$) and $k = \frac {-b}{2a} - \frac {\sqrt{b^2 - 4ac}}{2a}i$

Notice $z$ and $k$ are conjugates; Their real parts are equal and their imaginary parts are negatives of each other.

!!!!SO!!!!

IF $(x - z)(x -k) = ax^2 + by + c$ is a polynomial with only real coeficients and no real solutions, then z and k are conjugate complex numbers.

WE ARE HALF-WAY DONE!

!!!!NOW!!!! let's suppose z and k are conjugates. In other words, if $z = a + bi$ and $k = a - bi$, !!!THEN!!! z and k are the roots of the polynomial $(x -z)(x -k$. That polynomial is $(x - z)(x - k) = x^2 - (z + k)x + zk = x^2 - [(a + bi) + (a - bi)]x + (a + bi)(a - bi) = x^2 - 2ax + (a^2 - b^2)$ has coefficients, $1, 2a,$ and $(a^2 - b^2)$. These coefficients are all real numbers.

!!!SO!!!! IF x and z are conjugates THEN they are roots of a polynomial with real coefficients.

Now we have completed everything.

$z,$ and $k$ are complex roots to a second degree polynomial that has only real coefficients IF AND ONLY IF $z$ and $k$ are conjugates.

Answer 4

Well, if

$z = \bar k, \tag{1}$

then taking

$k = k_R + ik_I \tag{2}$

we have

$z = k_R - ik_I; \tag{3}$

thus

$k + z = 2k_R \tag{4}$

and

$kz = k_R^2 + k_I^2, \tag{5}$

both real; thus the implication

$z = \bar k \Rightarrow x^2 - (k + z)x + kz \in \Bbb R[x] \tag{6}$

is established.

Going the other way, if

$x^2 - (k + z)x + kz \in \Bbb R[x], \tag{7}$

writing

$z = z_R + iz_I, \tag{8}$

we have

$(k_R + z_R) + i(k_I + z_I) \in \Bbb R, \tag{9}$

whence

$k_I + z_I = 0, \tag{10}$

whence

$z_I = -k_I; \tag{11}$

then

$k_R z_R + k_I^2 + i k_I(z_R - k_R)$ $= (k_R + ik_I)(z_R - ik_I) = kz \in \Bbb R; \tag{12}$

thus

$k_I(z_R - k_R) = 0; \tag{13}$

since

$k_I \ne 0, \tag{14}$

$z_R = k_R \tag{15}$

and we are done; $z = \bar k$. QED.