Proving the Relationship between Complex Numbers and Trigonometric Functions

[loy]

Homework Statement

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

Homework Equations

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 don't know how to proceed the question... anyone can help me?

 

[HallsofIvy]

Homework Statement

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

Homework Equations

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 don't know how to proceed the question... anyone can help me?

 

[loy]

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

 

[SammyS]

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 .

 

[loy]

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

 

[SammyS]

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.

 

[loy]

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

 

[SammyS]

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 .

 

[loy]

thanks dude , i found the answer!

 

[foufou]

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]

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.

    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 .

 

[foufou]

yeah i know sin2(θ) + cos2(θ) = 1 .

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

 

[Munzi5]

It should be 6 + 3 Cosθ - i 3 Sinθ.
3 is missing in i SinθAnyway, I kept getting 9 / (5+4Cosθ) for (x + iy)(x - iy)

 

[loy]

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.

 

[Munzi5]

Head bang.

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

 

[loy]

welcome dude !