What is the angle made by a line with the real axis, given certain conditions?

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In summary, 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.
  • #1
utkarshakash
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Homework Statement


Consider a line [itex]\alpha\overline{z}+\overline{\alpha}z+i\beta [/itex] such that [itex]\frac{-\alpha}{\overline{\alpha}} = \lambda(1+i), \lambda \in R^{+}[/itex], then the angle made by line with real axis is [itex]\dfrac{\pi}{k},[/itex] then k is


Homework Equations



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

[itex] (-\overline{\alpha}\lambda(1+i) )\overline{z}+\overline{\alpha}z+i\beta[/itex].

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
[itex]z\overline{\alpha}-\overline{z}\alpha+b[/itex] (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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  • #2
utkarshakash said:

Homework Statement


Consider a line [itex]\alpha\overline{z}+\overline{\alpha}z+i\beta [/itex] such that [itex]\frac{-\alpha}{\overline{\alpha}} = \lambda(1+i), \lambda \in R^{+}[/itex], then the angle made by line with real axis is [itex]\dfrac{\pi}{k},[/itex] then k is
... then k is what? Is some information missing here?
utkarshakash said:

Homework Equations



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

[itex] (-\overline{\alpha}\lambda(1+i) )\overline{z}+\overline{\alpha}z+i\beta[/itex].

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
[itex]z\overline{\alpha}-\overline{z}\alpha+b[/itex] (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.
 
  • #3
Mark44 said:
... 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.
 
  • #4
utkarshakash said:
Consider a line [itex]\alpha\overline{z}+\overline{\alpha}z+i\beta [/itex]
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
haruspex said:
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

[itex]2(x_{1}x_{2}+y_1y_2) + i\beta=0 [/itex]
 
  • #6
utkarshakash said:
Consider a line [itex]\alpha\overline{z}+\overline{\alpha}z+i\beta [/itex]

haruspex said:
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.
 
  • #7
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:
  • #8
Mark44 said:
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.
 
  • #9
haruspex said:
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?
 
  • #10
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.
[itex]|\frac{-\alpha}{\overline{\alpha}}| = 1[/itex], 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
haruspex said:
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.
[itex]|\frac{-\alpha}{\overline{\alpha}}| = 1[/itex], 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
 

What is the angle made by a line?

The angle made by a line is the measure of the amount of rotation required to bring one line into the same position as another line.

How do you measure the angle made by a line?

The angle made by a line is typically measured in degrees using a protractor or other measuring tool.

What is the difference between acute, obtuse, and right angles?

An acute angle is less than 90 degrees, an obtuse angle is greater than 90 degrees, and a right angle is exactly 90 degrees.

Can the angle made by a line be negative?

No, the angle made by a line cannot be negative as it is always measured in a positive direction.

How can the angle made by a line affect other geometric shapes?

The angle made by a line can affect the overall shape and measurements of other geometric shapes, as it is a key factor in determining the relationships between different lines and angles.

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