# Angle made by the line

Gold Member
1. Homework Statement
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. Homework 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?

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Mark44
Mentor
1. Homework Statement
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
... then k is what? Is some information missing here?
2. Homework 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?
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.

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

haruspex
Homework Helper
Gold Member
2018 Award
Consider a line $\alpha\overline{z}+\overline{\alpha}z+i\beta$
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.

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

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

Mark44
Mentor
Consider a line $\alpha\overline{z}+\overline{\alpha}z+i\beta$
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.
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.

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.

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haruspex
Homework Helper
Gold Member
2018 Award
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.
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.

Gold Member
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.
So does that mean that the question is incorrect?

haruspex
Homework Helper
Gold Member
2018 Award
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?

Gold Member
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?

My teacher says its incorrect