Homework Help: When is the expression real

1. May 8, 2015

Bashyboy

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. 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.

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

2. May 8, 2015

PeroK

Not very impressive. Maybe the answer is pretty obvious.

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

Last edited: May 8, 2015
3. May 8, 2015

vela

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

4. May 8, 2015

Bashyboy

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

5. May 8, 2015

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.

6. May 8, 2015

PeroK

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

7. May 8, 2015

Bashyboy

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

8. May 8, 2015

PeroK

Complex conjugates have that property!

9. May 8, 2015

Bashyboy

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

10. May 8, 2015

PeroK

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

11. May 8, 2015

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.

12. May 8, 2015

PeroK

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

13. May 8, 2015

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?

14. May 8, 2015

PeroK

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 $\beta = i$.

15. May 8, 2015

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.

16. May 8, 2015

PeroK

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.

17. May 8, 2015

jbunniii

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

18. May 8, 2015

Ray Vickson

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?

Last edited: May 8, 2015