- #1

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(ix)(1+iy)=(3x+i4)/(x+3y)

I have tried grouping the equation into two equal complex numbers but have failed to find a solution which isolates x and y from the two quite long polynomial equations.

Does anybody know how to do this.

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- Thread starter anjunawoop
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- #1

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(ix)(1+iy)=(3x+i4)/(x+3y)

I have tried grouping the equation into two equal complex numbers but have failed to find a solution which isolates x and y from the two quite long polynomial equations.

Does anybody know how to do this.

- #2

vela

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Your approach sounds fine. Show us what you did so far.

- #3

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(xi)/(1 + yi) = (3x + 4i)/(x + 3y)

First rationalized the left hand fraction.

(xi)/(1 + yi) * (1 - yi)/(1 - yi) = (xi -yxi^2)/(1 + y^2)

= (yx + xi)/(1 + y^2)

Grouped real and imaginary terms

[ yx/(1 + y^2) ] + [ x/(1 + y^2) ]i

Then grouped real and imaginary terms on the right hand fraction

(3x + 4i)/(x + 3y) = [ 3x/(x + 3y) ] + [ 4/(x + 3y) ]i

Set the two complex numbers equal

[ yx/(1 + y^2) ] + [ x/(1 +y^2) ]i = [ 3x/(x + 3y) ] + [ 4/(x +3y) ]i

Take the two imaginary parts to be identical

Take the two real parts to be identical

[ yx/(1 + y^2) ] = [ 3x/(x +3y) ]

[ x/(1 + y^2) ] = [ 4/(x + 3y) ]

Expressing both fractions as polynomials

(yx)(x + 3y) = (3x)(1 + y^2)

(x)(x + 3y) = (4)(1 + y^2)

Then multiplying out both equations in x and y

1. yx^2 + 3y^2x = 3x + 3xy^2

2. x^2 + 3yx = 4 +4y^2

Grouping terms on one side to form two equations that equal zero.

1. yx^2 + 3y^2x - 3x - 3xy^2 = 0

2. x^2 +3yx - 4 - 4y^2 = 0

Now trying to find some way of dealing with these terms.

First rationalized the left hand fraction.

(xi)/(1 + yi) * (1 - yi)/(1 - yi) = (xi -yxi^2)/(1 + y^2)

= (yx + xi)/(1 + y^2)

Grouped real and imaginary terms

[ yx/(1 + y^2) ] + [ x/(1 + y^2) ]i

Then grouped real and imaginary terms on the right hand fraction

(3x + 4i)/(x + 3y) = [ 3x/(x + 3y) ] + [ 4/(x + 3y) ]i

Set the two complex numbers equal

[ yx/(1 + y^2) ] + [ x/(1 +y^2) ]i = [ 3x/(x + 3y) ] + [ 4/(x +3y) ]i

Take the two imaginary parts to be identical

Take the two real parts to be identical

[ yx/(1 + y^2) ] = [ 3x/(x +3y) ]

[ x/(1 + y^2) ] = [ 4/(x + 3y) ]

Expressing both fractions as polynomials

(yx)(x + 3y) = (3x)(1 + y^2)

(x)(x + 3y) = (4)(1 + y^2)

Then multiplying out both equations in x and y

1. yx^2 + 3y^2x = 3x + 3xy^2

2. x^2 + 3yx = 4 +4y^2

Grouping terms on one side to form two equations that equal zero.

1. yx^2 + 3y^2x - 3x - 3xy^2 = 0

2. x^2 +3yx - 4 - 4y^2 = 0

Now trying to find some way of dealing with these terms.

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- #4

I like Serena

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(xi)/(1 + yi) = (3x + 4i)/(x + 3y)

(xi)(x + 3y) = (3x + 4i)(1 + yi)

x²i + 3xyi = 3x + 4i + 3xyi - 4y

x²i = 3x + 4i - 4y

(4y-3x) + (x² - 4)i = 0

That doesn't look so bad does it?

I suspect you made a mistake somewhere, but I haven't figured out where yet.

- #5

vela

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Note that you can write these two equations asExpressing both fractions as polynomials

(yx)(x + 3y) = (3x)(1 + y^2)

(x)(x + 3y) = (4)(1 + y^2)

\begin{align*}

\frac{x(x+3y)}{1+y^2} &= \frac{3x}{y} \\

\frac{x(x+3y)}{1+y^2} &= 4

\end{align*}So you have y=(3/4)x. You can plug use that to eliminate y in one of the equations and solve for x.

Or you can do it the way ILS did it.

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