Complex numbers problem

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  • #1
chwala
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Homework Statement


if ## x + iy## = ## \frac a {b+ cos ∅ + i sin ∅} ##
then show that
##(b^2-1)(x^2+y^2)+a^2 = 2abx##


Homework Equations




The Attempt at a Solution



i let ## ... ##x + iy = ## \ a(b+cos ∅ - i sin ∅)##/ ##(b + cos ∅)^2 + sin^2∅ ##.......got stuck here....

alternatively i let
## b + cos ∅ + i sin ∅## = ## a/(x +iy)## then it follows that
## b+ cos ∅ + i sin ∅## = ## a(x-iy)/(x^2 +y^2)##
## x^2 + y^2 = a(x-iy)/ (b+ cos ∅+ i sin ∅)##...advise stuck again...am i doing something wrong...
 
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  • #2
BvU
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Hello chwala,

If two complex numbers are equal, you have two equations:
real parts must be equal
imaginary parts must be equal

I don't see you making use of that explicitly ?
 
  • #3
chwala
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Hello chwala,

If two complex numbers are equal, you have two equations:
real parts must be equal
imaginary parts must be equal

I don't see you making use of that explicitly ?
ok, let me try that and see will post on Monday greetings from Malaysia...
 
  • #4
Also, you will need to multiply the right-hand side of the original equation by the complex conjugate in order to get rid of the imaginary part in the denominator.

As a simple example, note the following:

$$\frac {1}{a + ib} \cdot \frac {a - ib}{a - ib} = \frac {a-ib}{a^2 + iab - iab - i^2 b^2} = \frac{a - ib}{a^2 - i^2 b^2} = \frac{a - ib}{a^2 + b^2}$$
 
  • #5
chwala
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Hello chwala,

If two complex numbers are equal, you have two equations:
real parts must be equal
imaginary parts must be equal

I don't see you making use of that explicitly ?
Still having a problem even after equating them...
 
  • #6
Ray Vickson
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Still having a problem even after equating them...
Show us what you get; explain where and why you are having a problem. We cannot help if we don't know what you are doing.
 
  • #7
I find it intuitive to force the solution to have a real denominator which leads to separation.The rest is simply abstract thinking which is the fun part.
 
  • #8
chwala
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I find it intuitive to force the solution to have a real denominator which leads to separation.The rest is simply abstract thinking which is the fun part.
X- (a cos ¤)/((b+cos ¤)^2+sin^2¤) = 0 and y + (a sin ¤)/((b + cos ¤)^2 + sin^2 ¤)=0
On solving simultaneous
-y(a cos ¤)-(ax sin ¤)=0
Sorry I am posting from my Android phone... Am I correct?
 
  • #9
chwala
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Sorry s
X- (a cos ¤)/((b+cos ¤)^2+sin^2¤) = 0 and y + (a sin ¤)/((b + cos ¤)^2 + sin^2 ¤)=0
On solving simultaneous
-y(a cos ¤)-(ax sin ¤)=0
Sorry I am posting from my Android phone... Am I correct?
Sorry small correction

y(ab + a cos ¤)+ ax sin ¤ = 0...
 
  • #10
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Homework Statement


if x+iyx+iy x + iy = ab+cos∅+isin∅ab+cos∅+isin∅ \frac a {b+ cos ∅ + i sin ∅}
then show that
(b2−1)(x2+y2)+a2=2abx(b2−1)(x2+y2)+a2=2abx(b^2-1)(x^2+y^2)+a^2 = 2abx
If two complex numbers are equal then their modulus squares are also equal. The modulus of square of LHS = x² + y² which occurs in the expression to be proved. So I suggest rationalize the denominator of RHS and write it as A + iB. and then the modulus squares of the two sides and simplify.Hope you get the result. But do something.

The difficulty with your work is that you are not able to get an equation or equations which has real numbers only. We all are suggesting you that only and you are doing nothingor not rpoerting your work in this direction or as per our advices.
 
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  • #11
chwala
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Thanks, I know modulus can solve... I am looking for another method...
 
  • #12
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Thanks, I know modulus can solve... I am looking for another method...
That is a good pursuit. Why that when you know that the the two representations are equivalent and even the two methods will also be equivalent. By other did you get the value of x and y in terms of real numbers a, b and phi?
 
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  • #13
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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) add to this what is given and you can get x immediately. Pursue this method as another method.
 
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  • #14
chwala
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Thanks, I know modulus can solve... I am looking for another method...
I will post the last steps in my modulus method tomorrow to show understanding and mastery, my concern is using another method apart from modulus..
 
  • #15
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my concern is using another method apart from modulus
I don't think there is another way. The right side of the original equation is a quotient with a complex number in the denominator. The only way I know to rewrite such a quotient in the form a + bi is to multiply the fraction by 1 in the form of the complex conjugate of the denominator over itself. The right side of the equation has to be written in the form of a + bi so as to equate its real and imaginary parts with those of the other side of the equation.
 
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  • #16
chwala
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I don't think there is another way. The right side of the original equation is a quotient with a complex number in the denominator. The only way I know to rewrite such a quotient in the form a + bi is to multiply the fraction by 1 in the form of the complex conjugate of the denominator over itself. The right side of the equation has to be written in the form of a + bi so as to equate its real and imaginary parts with those of the other side of the equation.
There is another way, without use of conjugate or modulus, let me post it now
 
  • #17
chwala
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## (x +iy)(b + cos Φ + i sin Φ) = a##
## xb + x cos Φ - y sin Φ = a ##
and
## yb + xsin Φ + y cos Φ = 0##
##x cos Φ - y sin Φ = a -xb##
## x sin Φ + y cos Φ = -yb##
on squaring each side of the equation,
##(x cos Φ - ysin Φ)^2 = (a-xb)^2##
##x^2cos^2Φ-2xycosΦ sin Φ+y^2sin^2Φ=a^2-2abx +x^2b^2##...............1
##(x sin Φ +y cosΦ)^2=(-yb)^2##
##x^2sin^2Φ+2xy sin Φ cos Φ+y^2cos^2Φ= y^2b^2##............................2
adding 1 and 2,
##x^2(cos^Φ+sin^2Φ)+y^2(sin^2Φ+cos^2Φ)= a^2-2abx+x^2b^2+y^2b^2##
## x^2+y^2= a^2-2abx +x^2b^2 + y^2b^2##
## 2abx = a^2 +x^2(-1+b^2) + y^2(-1+b^2)##
therefore,
##2abx = (b^2-1)(x^2+y^2)+ a^2##

I will post the modulus method later, so we can't use the conjugate method in this problem, still wondering why, i will give it another try...thanks guys...
 
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  • #18
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I will post the modulus method later, so we can't use the conjugate method in this problem, still wondering why, i will give it another try...thanks guys...
(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
 
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  • #19
Dick
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## (x +iy)(b + cos Φ + i sin Φ) = a##
## xb + x cos Φ - y sin Φ = a ##
and
## yb + xsin Φ + y cos Φ = 0##
##x cos Φ - y sin Φ = a -xb##
## x sin Φ + y cos Φ = -yb##
on squaring each side of the equation,
##(x cos Φ - ysin Φ)^2 = (a-xb)^2##
##x^2cos^2Φ-2xycosΦ sin Φ+y^2sin^2Φ=a^2-2abx +x^2b^2##...............1
##(x sin Φ +y cosΦ)^2=(-yb)^2##
##x^2sin^2Φ+2xy sin Φ cos Φ+y^2cos^2Φ= y^2b^2##............................2
adding 1 and 2,
##x^2(cos^Φ+sin^2Φ)+y^2(sin^2Φ+cos^2Φ)= a^2-2abx+x^2b^2+y^2b^2##
## x^2+y^2= a^2-2abx +x^2b^2 + y^2b^2##
## 2abx = a^2 +x^2(-1+b^2) + y^2(-1+b^2)##
therefore,
##2abx = (b^2-1)(x^2+y^2)+ a^2##
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)##.
 
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  • #20
BvU
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if the given relation is true for any ϕ
It clearly is not. So the qualification 'vacuously' is unfounded.
I think Chwala did well.

a tip for Chwala: you want to use ## \sin\phi ## to get ##\ \sin\phi \ ## instead of ## sin\phi ## ( ##sin\phi \ ##)
 
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  • #21
## (x +iy)(b + cos Φ + i sin Φ) = a##
## xb + x cos Φ - y sin Φ = a ##
and
## yb + xsin Φ + y cos Φ = 0##
##x cos Φ - y sin Φ = a -xb##
## x sin Φ + y cos Φ = -yb##
on squaring each side of the equation,
##(x cos Φ - ysin Φ)^2 = (a-xb)^2##
##x^2cos^2Φ-2xycosΦ sin Φ+y^2sin^2Φ=a^2-2abx +x^2b^2##...............1
##(x sin Φ +y cosΦ)^2=(-yb)^2##
##x^2sin^2Φ+2xy sin Φ cos Φ+y^2cos^2Φ= y^2b^2##............................2
adding 1 and 2,
##x^2(cos^Φ+sin^2Φ)+y^2(sin^2Φ+cos^2Φ)= a^2-2abx+x^2b^2+y^2b^2##
## x^2+y^2= a^2-2abx +x^2b^2 + y^2b^2##
## 2abx = a^2 +x^2(-1+b^2) + y^2(-1+b^2)##
therefore,
##2abx = (b^2-1)(x^2+y^2)+ a^2##

I will post the modulus method later, so we can't use the conjugate method in this problem, still wondering why, i will give it another try...thanks guys...
I did it another way and spent two ****en days on this.Thankyou for such a question !!!!!
 
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  • #22
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Ok, I don't get this. You seem to have eliminated ϕϕ\phi to get a relation between the real numbers xxx, yyy, aaa and bbb. But you can get a stronger conclusion if the given relation is true for any ϕϕ\phi. Putting ϕ=0ϕ=0\phi =0 and ϕ=πϕ=π\phi = \pi gives you x+iy=ab+1=ab−1x+iy=ab+1=ab−1x+iy=\frac{a}{b+1}=\frac{a}{b-1} leading to the conclusion a=x=y=0a=x=y=0a=x=y=0 and bbb can be anything. Which means your derived relation works, but kind of vacuously. I think what you probably mean is that xxx and yyy are functions of ϕϕ\phi. In which case it's probably better to write them as x(ϕ)x(ϕ)x(\phi) and y(ϕ)y(ϕ)y(\phi).
I do not agree. when a complex number x+iy has been given equal to another complex containing a, b and ϕ, then it is obvious that both x and y will be function of the variables a, b and ϕ, although they may not be independent variables.

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.

Of course the problem has become very interesting as a, b and ϕ cannot be independent one can try to express ϕ in terms of a and b and then find the domain of ϕ over which the given equality holds good.

Sir, as you have not got the solution I will also solve it completely by method and find some more clarity or conclusion.
 
  • #23
It doesn't....
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
(R-I)2 = R2+I2

x2 + y2 = a2/(R+I)2 = (x+yi)^2

ax = a2R/((R+I)(R-I))
a2 = (x2 + y2 )((R+I)2)

2bax = 2bR(x2 + y2 )((R+I)2)/((R+I)(R-I))
∴ prove
(b2-1)(x2+y2 ) + (x2 + y2 )((R+I)2) = 2bR(x2 + y2 )((R+I)2)/((R+I)(R-I))

(b2-1)+ ((R+I)2) = 2b(R(R+I)2)/((R+I)(R-I))

((b2-1)+ ((R+I)2))(R-I) = 2bR(R+I)

RHS
Sub R and I in terms of non R AND I terms
2b3+2bcos2(m) +4bcos2(m) + (2b2sin(m)+2bsin(m)cos(m))i

LHS
becomes
(-b+2b3+bcos(2m)+2bcos2 (m) +2bsin2 (m) +4b2 cos(2m) )+ (2sin(m)+bsin(2m))I
relate them both

(2b2sin(m)+2bsin(m)cos(m))i = (2sin(m)+bsin(2m))i
2b2sin(m)+2bsin(m)cos(m)= 2bsin(m)cos(m)+2sin(m)
2b2sin(m) = 2sin(m)
this only works if m = 0 or π
Which would suit the solution
hence no.
 
  • #24
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x2 + y2 = a2/(R+I)2 = (x+yi)^2
This step is absolutely incorrect x^2 + y^2 cannot be equal to (x+iy)^2 unless y = 0
 
  • #25
This step is absolutely incorrect x^2 + y^2 cannot be equal to (x+iy)^2 unless y = 0
In this case it does, show me where my math was specifically wrong.
also
x^2 +y^2 = x^2-y^2+2xiy
2y^2=2xiy
y=xi or x = 0
 

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