Can Complex Variables Always Yield a Real Expression?

[Bashyboy]

Homework Statement

I have the expression $a_2\overline{β} + a_3β$ involving complex variables, where $|β|≤1$. I was wondering, is it possible to determine for what $a_2$ and $a_3$ the expression is always real, for every $β$ satisfying $|β|≤1$; or is there insufficient information?

Homework Equations

The Attempt at a Solution

I was under the impression an inquiry constituted an attempt. At any rate, I am not asking anyone to solve the problem; I am merely asking if solving such a problem is possible given the information. This doesn't come from a textbook, although I am sure it is considered a standard, textbook question.

How is this for an attempt?

$a_2 \overline{\beta} + a_3 \beta = r$, where $r$ is some real number.

$a_2 = \frac{r - a_3 \beta}{\overline{\beta}} = e^{i \theta} - a_3 e^{2 i \theta}$

 

[PeroK]

Homework Statement

I have the expression $a_2\overline{β} + a_3β$ involving complex variables, where $|β|≤1$. I was wondering, is it possible to determine for what $a_2$ and $a_3$ the expression is always real, for every $β$ satisfying $|β|≤1$; or is there insufficient information?

Homework Equations

The Attempt at a Solution

I was under the impression an inquiry constituted an attempt. At any rate, I am not asking anyone to solve the problem; I am merely asking if solving such a problem is possible given the information. This doesn't come from a textbook, although I am sure it is considered a standard, textbook question.

How is this for an attempt?

$a_2 \overline{\beta} + a_3 \beta = r$, where $r$ is some real number.

$a_2 = \frac{r - a_3 \beta}{\overline{\beta}} = e^{i \theta} - a_3 e^{2 i \theta}$

Not very impressive. Maybe the answer is pretty obvious.

Edit: when is the sum of two complex numbers real?

 

[vela]

Try writing $\beta = x + iy$ and then require the imaginary part of the expression you wrote to equal 0.

 

[Bashyboy]

Wouldn't I have to do the same for $a_2$ and $a_3$, seeing as I am trying to solve for these?

 

[Bashyboy]

As PeroK noted, my situation is a particular case of a more general result--namely, that $z_1 + z_2 = 0 \implies Im~z_2 = -Im~z_1$. But this doesn't seem particularly illuminating as $Im~(a_2 \overline{\beta}) = - Im~(a_3 \beta)$ produces a somewhat intricate equation.

 

[PeroK]

As PeroK noted, my situation is a particular case of a more general result--namely, that $z_1 + z_2 = 0 \implies Im~z_2 = -Im~z_1$. But this doesn't seem particularly illuminating as $Im~(a_2 \overline{\beta}) = - Im~(a_3 \beta)$ produces a somewhat intricate equation.

I would say that $Im(z_1) = -Im(z_2)$ is very illuminating. Can you think of when that is the case?

 

[Bashyboy]

When what is the case? When $Im~(z_1) = -Im~(z_2)$ is true? Isn't it true when $y_2 = -y_1$?

 

[PeroK]

When what is the case? When $Im~(z_1) = -Im~(z_2)$ is true? Isn't it true when $y_2 = -y_1$?

Complex conjugates have that property!

 

[Bashyboy]

I don't think we can claim that $z_1$ and $z_2$ are complex conjugates as they could have different real parts.

 

[PeroK]

I don't think we can claim that $z_1$ and $z_2$ are complex conjugates as they could have different real parts.

No, but IF they are complex conjugates, then they have that property.

 

[Bashyboy]

Sure, I would agree with that, but I can't conclude from that that $a_2 \overline{\beta}$ and $a_3 \beta$ are complex conjugates, if that's what you are aiming at.

 

[PeroK]

Sure, I would agree with that, but I can't conclude from that that $a_2 \overline{\beta}$ and $a_3 \beta$ are complex conjugates, if that's what you are aiming at.

Not directly, no. But if $a_2$ and $a_3$ are complex conjugates?

 

[Bashyboy]

Yes, I agree. But I am trying to deduce what $a_2$ and $a_3$ are, not assign a property to them. My problem is, how do I not know there are more cases?

 

[PeroK]

Yes, I agree. But I am trying to deduce what $a_2$ and $a_3$ are, not assign a property to them. My problem is, how do I not know there are more cases?

That's the next question. You know that if $a_2$ and $a_3$ are conjugates, then $a_2 \bar{\beta} + a_3 \beta$ is real for all $\beta$.

Now you have to show that they must be conjugates. Hint: consider $\beta = i$.

 

[jbunniii]

$\alpha_2$ and $\alpha_3$ need not be conjugates. A complex number is real if and only if it equals its own conjugate, so the requirement that $\alpha_2 \overline{\beta} + \alpha_3 \beta$ is real is equivalent to
$$\alpha_2 \overline{\beta} + \alpha_3 \beta = \overline{\alpha_2} \beta + \overline{\alpha_3} \overline{\beta}$$
Rearranging, this is the same as
$$\beta(\alpha_3 - \overline{\alpha_2}) = \overline{\beta}(\overline{\alpha_3} - \alpha_2)$$
Writing $\alpha_3 = a + bi$ and $\alpha_2 = c + di$, the above becomes
$$\beta(a - c + i(b - d)) = \overline{\beta}(a - c - i(b + d))$$
This is one linear equation with four unknowns, so there are three degrees of freedom in specifying $a,b,c,d$: you can set any three of them freely, and solve for the fourth. If it was necessary for $\alpha_2$ and $\alpha_3$ to be conjugates, you would only have two degrees of freedom available.

 

[PeroK]

$\alpha_2$ and $\alpha_3$ need not be conjugates. A complex number is real if and only if it equals its own conjugate, so the requirement that $\alpha_2 \overline{\beta} + \alpha_3 \beta$ is real is equivalent to
$$\alpha_2 \overline{\beta} + \alpha_3 \beta = \overline{\alpha_2} \beta + \overline{\alpha_3} \overline{\beta}$$
Rearranging, this is the same as
$$\beta(\alpha_3 - \overline{\alpha_2}) = \overline{\beta}(\overline{\alpha_3} - \alpha_2)$$
Writing $\alpha_3 = a + bi$ and $\alpha_2 = c + di$, the above becomes
$$\beta(a - c + i(b - d)) = \overline{\beta}(a - c - i(b + d))$$
This is one linear equation with four unknowns, so there are three degrees of freedom in specifying $a,b,c,d$: you can set any three of them freely, and solve for the fourth. If it was necessary for $\alpha_2$ and $\alpha_3$ to be conjugates, you would only have two degrees of freedom available.

Well, except this was supposed to be the case for all $\beta$. If the expression is real for all $\beta$ then $a_2$ and $a_3$ must be conjugates.

 

[jbunniii]

Well, except this was supposed to be the case for all $\beta$. If the expression is real for all $\beta$ then $a_2$ and $a_3$ must be conjugates.

Oops, sorry, I missed that! In that case, I agree: consider $\beta = i$.

 

[Ray Vickson]

$\alpha_2$ and $\alpha_3$ need not be conjugates. A complex number is real if and only if it equals its own conjugate, so the requirement that $\alpha_2 \overline{\beta} + \alpha_3 \beta$ is real is equivalent to
$$\alpha_2 \overline{\beta} + \alpha_3 \beta = \overline{\alpha_2} \beta + \overline{\alpha_3} \overline{\beta}$$
Rearranging, this is the same as
$$\beta(\alpha_3 - \overline{\alpha_2}) = \overline{\beta}(\overline{\alpha_3} - \alpha_2)$$
Writing $\alpha_3 = a + bi$ and $\alpha_2 = c + di$, the above becomes
$$\beta(a - c + i(b - d)) = \overline{\beta}(a - c - i(b + d))$$
This is one linear equation with four unknowns, so there are three degrees of freedom in specifying $a,b,c,d$: you can set any three of them freely, and solve for the fourth. If it was necessary for $\alpha_2$ and $\alpha_3$ to be conjugates, you would only have two degrees of freedom available.

If we write $v = a_3 - \bar{a_2}$ your condition is that $\beta v = \overline{\beta v}$, so $\beta v =$ real for all $|\beta| \leq 1$. Write $\beta = r e^{i t}$ and $v = R e^{i w}$. If $rR \neq 0$ this implies $e^{i(t+w)} =$ real for all $w \in [0, 2 \pi)$, so $\sin(t+w) = 0$ for all $t$. Do you think that is possible?