# Angle made by the line

1. Oct 26, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
Consider a line $\alpha\overline{z}+\overline{\alpha}z+i\beta$ such that $\frac{-\alpha}{\overline{\alpha}} = \lambda(1+i), \lambda \in R^{+}$, then the angle made by line with real axis is $\dfrac{\pi}{k},$ then k is

2. Relevant equations

3. The attempt at a solution
I have tried nearly every possible method but still did not get the answer. Suppose I substitute α in the equation of given line I get

$(-\overline{\alpha}\lambda(1+i) )\overline{z}+\overline{\alpha}z+i\beta$.

But I don't see any point in doing these things as it won't help me.

Second Method
The equation of line perpendicular to this line is given by
$z\overline{\alpha}-\overline{z}\alpha+b$ (for some 'b')
If somehow I could get the slope of this line I would get my answer. But the problem is how?

2. Oct 26, 2012

### Staff: Mentor

... then k is what? Is some information missing here?
I haven't worked this all the way through, but I started by writing $\alpha = \alpha_1 + \alpha_2 i$ and z = z1 + z2 i to write the given complex number in rectangular form.

I'm assuming that $\beta$ is also complex.

3. Oct 26, 2012

### utkarshakash

In this question I have to find k.
Here β is not complex. It is real.

4. Oct 27, 2012

### haruspex

How does that define a line? Do you mean this expression = 0?
If so, I think there must be something wrong. Substitute z=x+iy and similarly for alpha and notice what happens to the imaginary part.

5. Oct 27, 2012

### utkarshakash

Of course, that expression is equal to 0.
Doing so I get the following

$2(x_{1}x_{2}+y_1y_2) + i\beta=0$

6. Oct 27, 2012

### Staff: Mentor

IMO the problem was stated unclearly. I interpret the problem this way: Consider the [STRIKE]line[/STRIKE] complex number ...

Graphing the complex number in the plane produces a line segment that makes a certain angle with the positive real axis.

7. Oct 27, 2012

### aralbrec

The equation relating α and α* (conjugate) places a constraint on what values α can take on as a function of (real) lambda. For example, suppose α is written in polar form, what does that constraint say about α? You'll find that λ is fixed and α is arbitrary length at a fixed angle.

But there is something wrong with that equation. You can see there is trouble because αz* and α*z are complex conjugates of each other so that their sum is real. So your final equation is a+jβ = 0 for some real a, which can only be satisfied if a and β are both zero. ie any solutions (if there is more than one) will lie on the real axis.

Last edited: Oct 27, 2012
8. Oct 27, 2012

### haruspex

I thought of that, but I still don't see it. What is being varied to generate the complex number? Presumably it's z, which I take to be a complex number, so it could generate the whole plane, not just a line. As it happens, the imaginary parts of the first two terms cancel, so it ends up generating the line x+iβ. But that makes nonsense of the rest of the question.

9. Oct 28, 2012

### utkarshakash

So does that mean that the question is incorrect?

10. Oct 28, 2012

### haruspex

I've thought of another interpretation: it's the lambda that's the independent variable. Varying that, keeping z and beta constant, looks like it should make the expression generate a line. But that runs into another problem.
$|\frac{-\alpha}{\overline{\alpha}}| = 1$, so |λ| = 1/√2, leaving only two possible values for λ.
I'm not certain the question is wrong, but I can't make sense of it. Have you triple-checked it's copied out correctly?

11. Oct 29, 2012

### utkarshakash

My teacher says its incorrect