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Angle made by the line

utkarshakash
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1. 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


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

[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?
 

Answers and Replies

32,926
4,627
1. 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?
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

[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.
 
utkarshakash
Gold Member
855
13
... 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
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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.
 
utkarshakash
Gold Member
855
13
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]
 
32,926
4,627
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.
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.
 
296
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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:
haruspex
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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.
 
utkarshakash
Gold Member
855
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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
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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?
 
utkarshakash
Gold Member
855
13
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
 

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