# Proving of complex number

## Homework Statement

Given that, x+yi=3/[2+cos θ+i(sin θ)] , prove that x^2+y^2=4x-3

r^2=x^2+y^2

## The Attempt at a Solution

since we know that x=r sinθ , y=r cosθ,
i multiply r on the right hand side to equal 3r/[2r+x+yi], then multiply its comjugate (2r+x)-yi to equal (3rx+6r^2 -3ryi) / [(2r+x)^2 + y^2] . then i equal the imaginary part on left handside, yi= -3ryi/[(2r+x)^2 + y^2],
(2r+x)^2 + y^2 = -3r,
4r^2+x^2+ 4rx +y^2 = -3r, from the Relevant equations,
5r^2+4rx= -3r ,divide both sides by r
5r=-4x-3,
r=-(4x+3)/5,
x^2+y^2= [-(4x+3)/5]^2, until here, i dont know how to proceed the question... anyone can help me?

HallsofIvy
Homework Helper

## Homework Statement

Given that, x+yi=3/[2+cos θ+i(sin θ)] , prove that x^2+y^2=4x-3

r^2=x^2+y^2

## The Attempt at a Solution

since we know that x=r sinθ , y=r cosθ,
We know that if θ is the angle the line from 0 to x+ yi makes with the positive real axis. Is that given? You don't mention it in the statement of the problem.

If that is true then the right side is simply 3/(2+ x+ iy) and it would seem reasonable to multiply both sides by 2+ x+ iy giving (x+ iy)(2+ x+ iy)= 2x+ 2iy+ x^2+ ixy+ ixy- y^2= 3. That is NOT, in general, x^2+ y^2= 4x- 3.
i multiply r on the right hand side to equal 3r/[2r+x+yi], then multiply its comjugate (2r+x)-yi to equal (3rx+6r^2 -3ryi) / [(2r+x)^2 + y^2] . then i equal the imaginary part on left handside, yi= -3ryi/[(2r+x)^2 + y^2],
(2r+x)^2 + y^2 = -3r,
4r^2+x^2+ 4rx +y^2 = -3r, from the Relevant equations,
5r^2+4rx= -3r ,divide both sides by r
5r=-4x-3,
r=-(4x+3)/5,
x^2+y^2= [-(4x+3)/5]^2, until here, i dont know how to proceed the question... anyone can help me?

not given , my lecturer just gave me the question as i've written above.
So, what should I assume?

SammyS
Staff Emeritus
Homework Helper
Gold Member
not given , my lecturer just gave me the question as i've written above.
So, what should I assume?
But you did assume that x=r sinθ and y=r cosθ , so what Halls said about θ is true.

An alternate way to solve this problem, with no assumptions is to rationalize the denominator of $\displaystyle \ \ \frac{3}{2+\cos(\theta)+i\sin(\theta)}\ \$ and identify x and y by equating real parts and equating imaginary parts.

Then compare x2 + y2 with 4x-3 .

how to do equating the parts? they are cos and sin there.

SammyS
Staff Emeritus
Homework Helper
Gold Member
how to do equating the parts? they are cos and sin there.
Rationalize the denominator of $\displaystyle \ \ \frac{3}{2+\cos(\theta)+i\sin(\theta)}\ \$ to determine what is the real part and what is the imaginary part.

Since that's equal to x + iy, then you have x is equal to the real part, and y is equal to the imaginary part.

you mean multiply the conjugate of 2+cos(θ)+isin(θ) in order to find the x and y?

SammyS
Staff Emeritus
Homework Helper
Gold Member
you mean multiply the conjugate of 2+cos(θ)+isin(θ) in order to find the x and y?
Just so there's no misinterpretation, I mean:

Multiply $\displaystyle \ \ \frac{3}{2+\cos(\theta)+i\sin(\theta)}\ \$ by $\displaystyle \ \ \frac{2+\cos(\theta)-i\sin(\theta)}{2+\cos(\theta)-i\sin(\theta)}\ \$ in order to find the x and y .

thanks dude , i found the answer!

hey

I could not get the solution

I multiply and divide the RHS with its conjugate 2+cos∅-isin∅ which arrives at
6+3cos∅+isin∅ 6+3cos∅+isin∅
-------------------- = ---------------------------
(2+cos∅)^2 +sin^2∅ 4+4cos∅+cos^2∅+sin^2∅

How to equate the LHS

SammyS
Staff Emeritus
Homework Helper
Gold Member
hey

I could not get the solution

I multiply and divide the RHS with its conjugate 2+cos∅-isin∅ which arrives at
6+3cos∅+isin∅ 6+3cos∅+isin∅
-------------------- = ---------------------------
(2+cos∅)^2 +sin^2∅ 4+4cos∅+cos^2∅+sin^2∅

How to equate the LHS
Use [ code] [ /code] tags to write such fractions.

Code:
    6+3cosθ+isinθ                         6+3cosθ+isinθ
--------------------          =          ---------------------------
(2+cosθ)^2 +sin^2∅                      4+4cosθ+cos^2θ+sin^2θ

Don't forget, sin2(θ) + cos2(θ) = 1 .

yeah i know sin2(θ) + cos2(θ) = 1 .

but even then i'm stuck, can u help me further

It should be 6 + 3 Cosθ - i 3 Sinθ.
3 is missing in i Sinθ

Anyway, I kept getting 9 / (5+4Cosθ) for (x + iy)(x - iy)

Last edited:
the solution is like this
3/[2+cos θ+i(sin θ)]
= [3(2+Cosθ)- 3 Sinθ i] / [(2+cos θ)^2 + (sin θ)^2]

So, from LHS, we get
x=3(2+Cosθ)/ [(2+cos θ)^2 + (sin θ)^2] while
y=3 Sinθ/ [(2+cos θ)^2 + (sin θ)^2]

On LHS: x^2+y^2
=[9(2+Cosθ)^2+9Sinθ^2]/ [(2+cos θ)^2 + (sin θ)^2]^2
=9/[(2+cos θ)^2 + (sin θ)^2]
since the denominator and numerator has a common factor (2+cos θ)^2 + (sin θ)^2

On RHS: 4x-3
= 4*{3(2+Cosθ)/ [(2+cos θ)^2 + (sin θ)^2]} - 3
={24 + 12cosθ-3[(2+cos θ)^2 + (sin θ)^2]}/[(2+cos θ)^2 + (sin θ)^2]
={24+12cosθ-12-12cosθ-3cos2(θ)-3sin2(θ)}/[(2+cos θ)^2 + (sin θ)^2]
={24+12cosθ-12-12cosθ-3*(sin2(θ) + cos2(θ))}/[(2+cos θ)^2 + (sin θ)^2]
=9/[(2+cos θ)^2 + (sin θ)^2]

which is equal to the LHS
It is proven.

It did not strike that I can use L.H.S = R.H.S.

I tried to get 4x-3 from x^2 + y^2.
x^2+y^2 can be derived from (x + iy)(x-iy), then tried to simplify the x^2+y^2 to 4x-3.

Thank you very much Loy! Good day

Last edited:
welcome dude !