Complex numbers problem

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  • #26
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x2 + y2 = a2(R2 + I2)/((R+I)(R-I))2
mistake is in this step applying the formula (a+b)(a-b), where a and b can be complex also as the way you have defined I,I is complex, should be written as
R^2 - I^2 and not as R^2 + I^2. The way you have defined them R^2 >0 and I^2 <0
 
  • #27
chwala
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Ok, I don't get this. You seem to have eliminated ##\phi## to get a relation between the real numbers ##x##, ##y##, ##a## and ##b##. But you can get a stronger conclusion if the given relation is true for any ##\phi##. Putting ##\phi =0## and ##\phi = \pi## gives you ##x+iy=\frac{a}{b+1}=\frac{a}{b-1}## leading to the conclusion ##a=x=y=0## and ##b## can be anything. Which means your derived relation works, but kind of vacuously. I think what you probably mean is that ##x## and ##y## are functions of ##\phi##. In which case it's probably better to write them as ##x(\phi)## and ##y(\phi)##.
sorry just a small error, kindly note that
##cos^2 Φ + sin^2Φ = 1##
that is how i eliminated Φ.
 
  • #28
chwala
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i will post modulus method apart from use of complex conjugate and other method demonstrated on my thread on thursday.....i now have 3 ways of solving the problem
 
  • #29
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My second objection is by putting x = 0 and x = π, you are converting the problem into real domain so y will be obviously zero, so in the real domain may be this problem does not have solution.
I made an error here it should be φ = 0 and φ = π
 
  • #30
mistake is in this step applying the formula (a+b)(a-b), where a and b can be complex also as the way you have defined I,I is complex, should be written as
R^2 - I^2 and not as R^2 + I^2. The way you have defined them R^2 >0 and I^2 <0
y = -asin(x)/(R+I)(R-I)
square it...
y^2=a^2sin^2(x)/(R+I)^2(R-I)^2
= -a^2I^2/(R+I)^2(R-I)^2

oh didn't see that makes sense now!!!
will post the solution now :P
 
  • #31
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(x² +y²) (b² +1 +2bcosφ) = a² ---------------------- (3), we need to eliminate cosφ and bring in x
For that
Writing original expression
(x+iy) = [a/(b+cosφ+i sinφ)] ----------------------(4) Taking complex conjugate
(x-iy) = [a/(b+cosφ-i sinφ)] -----------------------(5); adding (4) and (5) we get
[2a(b+cosφ)/(b² +1 +2bcosφ)] = 2x --------------(6) multiplying (3) and (6) we get
[2a(b+cosφ)(x² +y²)] = 2xa² or
[2(b+cosφ)(x² +y²)] = 2xa ------ (7)
Rewrite (3) and (7) as
(b² +1 +2bcosφ) = a² /((x² +y²) ----------- (3') and rewritten (7) becomes, [2(b+cosφ)(x² +y²)] = 2xa ------ (7) or (b+cosφ) = xa/(x² +y²), multiplying by 2b gives
2b²+2bcos φ = 2bxa/(x² +y²), ------ (7'). Now (7') - (3') gives
(b² -1)(x² +y²) = 2abx - a² or
2abx = (b² -1)(x² +y²) + a²
Although I will not like to call them different methods but using complex conjugate property gives us the shortest, neatest, simplest and the sweetest derivation.
 
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  • #32
x+yi = a/(b+cis(m)) | R= b+cos(m) ,I = isin(m)
x+yi = a(R-I)/((R+I)(R-I))
= aR/((R+I)(R-I)) - aI/((R+I)(R-I))
x2+ y2 = a2(R2- I2)/((R+I)(R-I))2
a2=(x2+ y2)(R2- I2)
x= aR/(R2-I2)
2bax = 2ba2R/(R2-I2)
=2bR(x^2+y^2)
if
(b2-1)(x2+y2)+a2=2bR(x2+ y2)
(b2-1)(x2+y2)+(x2+ y2)(R2- I2)=2bR(x2+ y2)
b2-1+R2- I2=2bR
sub R and I
b2-1+cos2(m)+sin2(m)+b2+2bcos(m) = 2b2+2bcos(m)
2b2+2bcos(m) +1 - 1 = 2b2+2bcos(m)

Hence Q.E.D
 
  • #33
Ray Vickson
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y = -asin(x)/(R+I)(R-I)
square it...
y^2=a^2sin^2(x)/(R+I)^2(R-I)^2
= -a^2I^2/(R+I)^2(R-I)^2

oh didn't see that makes sense now!!!
will post the solution now :P
NO, do not post the solution. That is not allowed under PF rules! If YOU had posed the problem it would be OK for you to post the solution, but not if somebody else posted the problem. (It is also OK to post and alternative solution if the OP has already posted a valid solution.)
 
  • #34
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NO, do not post the solution. That is not allowed under PF rules! If YOU had posed the problem it would be OK for you to post the solution, but not if somebody else posted the problem. (It is also OK to post and alternative solution if the OP has already posted a valid solution.)
I don't think there's a problem here. The OP posted a solution many posts back.
 
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  • #35
Ray Vickson
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I don't think there's a problem here. The OP posted a solution many posts back.
I did allow for that possibility in the last sentence. I have not tried to follow all the interweaving of this convoluted thread.
 
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  • #36
I did allow for that possibility in the last sentence. I have not tried to follow all the interweaving of this convoluted thread.
Thankyou, didn't know you couldn't post solutions before OP.
 
  • #37
Ray Vickson
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I don't think there's a problem here. The OP posted a solution many posts back.
See post #36.
 
  • #38
chwala
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Thankyou, didn't know you couldn't post solutions before OP.
I posted a solution in post (17) and there was another solution in post (31). There is also another way of doing it using modulus, if given time i will post it here for research. In my opinion, solution (31) gives us the quickest and most friendliest of solutions.
 
  • #39
Ray Vickson
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I posted a solution in post (17) and there was another solution in post (31). There is also another way of doing it using modulus, if given time i will post it here for research. In my opinion, solution (31) gives us the quickest and most friendliest of solutions.
I think you are right, unless there is some "geometric" method that helped the person who proposed the problem to "invent" it. After all, how on earth would one ever guess that such a relationship could hold?
 
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  • #40
mathwonk
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i would start by setting the absolute values of both sides equal.
 
  • #41
chwala
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I think you are right, unless there is some "geometric" method that helped the person who proposed the problem to "invent" it. After all, how on earth would one ever guess that such a relationship could hold?
This is from the pure maths textbook by Prof. C. J Tranter, 1964 1st edition. i find the old maths texts quite intriguing in terms of approach to tackling problems.
 
  • #42
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The operation of complex conjugation is very powerful. We know that complex conjugate of sum or product or even division of two complex numbers is the sum, or product or division of their conjugates. So you can immediately write

x-iy = a/(b+cos φ -i sinφ) ....... 1 add to this what is given and you can get x immediately. Pursue this method as another method.
given x+iy = a/(b+cos φ +i sinφ)----- 2

Multiply 1 and 2
x^2 +y^2 = [(a^2 )/{(1+b^2+2b*cosφ)}]-------- 3
Add 1 and 2
2x = [{2a(b+cosφ)}/(1+b^2+2b*cosφ)] or
x = [{a(b+cosφ)}/(1+b^2+2b*cosφ)] ------------- 4
(x+iy)(b+cosφ+i sinφ) = a-----(1), taking complex conjugate on both sides we get
(x-iy)(b+cosφ-i sinφ) = a -------(2) multiplying LHSs and RHSs we get
(x² +y²) (b² +1 +2bcosφ) = a² ---------------------- (3), we need to eliminate cosφ and bring in x
For that
Writing original expression
(x+iy) = [a/(b+cosφ+i sinφ)] ----------------------(4) Taking complex conjugate
(x-iy) = [a/(b+cosφ-i sinφ)] -----------------------(5); adding (4) and (5) we get
[2a(b+cosφ)/(b² +1 +2bcosφ)] = 2x --------------(6) multiplying (3) and (6) we get
[2a(b+cosφ)(x² +y²)] = 2xa² or
[2(b+cosφ)(x² +y²)] = 2xa ------ (7)
Eliminating cosφ between (3) and (7) should give you the result.
They are all equivalent methods because you said complex conjugate will not work I did this all work
Hope it helps
Rewriting 3 gives us
(x² +y²) (b²-1 +1 +1 +2bcosφ) = a² or
{(b²-1) +2*(1 +bcosφ)] = a²/((x² +y²))
From 7 we have
(b+cosφ) =[ (xa)/(x² +y²)] ------ (7)multiplying by b gives
(b²+bcosφ) =[ (abx)/(x² +y²)] or
bcosφ =[ (xa)/(x² +y²)] - b² ---------- 8
Substituting 8 in 3 we get
(x² +y²) [(b² +1 +[ (2abx)/(x² +y²)] - 2b² ] = a² simplifying we get
(x² +y²) [(-b² +1 +[ (2abx)/(x² +y²)] = a² or
(x² +y²) (-b² +1) +2abx = a² or
2abx = a² + (x² +y²) (b² -1)

I think this was what was to be proved
 
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  • #43
chwala
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given x+iy = a/(b+cos φ +i sinφ)----- 2

Multiply 1 and 2
x^2 +y^2 = [(a^2 )/{(1+b^2+2b*cosφ)}]-------- 3
Add 1 and 2
2x = [{2a(b+cosφ)}/(1+b^2+2b*cosφ)] or
x = [{a(b+cosφ)}/(1+b^2+2b*cosφ)] ------------- 4

Rewriting 3 gives us
(x² +y²) (b²-1 +1 +1 +2bcosφ) = a² or
{(b²-1) +2*(1 +bcosφ)] = a²/((x² +y²))
From 7 we have
(b+cosφ) =[ (xa)/(x² +y²)] ------ (7)multiplying by b gives
(b²+bcosφ) =[ (abx)/(x² +y²)] or
bcosφ =[ (xa)/(x² +y²)] - b² ---------- 8
Substituting 8 in 3 we get
(x² +y²) [(b² +1 +[ (2abx)/(x² +y²)] - 2b² ] = a² simplifying we get
(x² +y²) [(-b² +1 +[ (2abx)/(x² +y²)] = a² or
(x² +y²) (-b² +1) +2abx = a² or
2abx = a² + (x² +y²) (b² -1)

I think this was what was to be proved
is this different from your post 31?
 
  • #44
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is this different from your post 31?
No. Just different elimination process.
 
  • #45
chwala
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ok ....
 

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