Complex numbers-finding real number pairs

**Neverquit** · March 19th 2010, 12:34 AM

Hello,
I am having trouble with this question:

"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

Any help would be appreciated.

Regards,
Neverquit

**Deadstar** · March 19th 2010, 03:39 AM

Multiply both sides by 1+pi

Then you get...

3 + 5i = (1 + pi)(q + 4i) = q + pqi + 4i - 4p

=> 3 + 5i = (q - 4p) + 4i + pqi

=> 3 + i = (q - 4p) + pqi

So you want to solve...

q - 4p = 3

pq = 1

**Neverquit** · March 19th 2010, 05:02 AM

I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.

There is also apparently another solution of 4, -1 which I can’t find.

How do I find it?

**Plato** · March 19th 2010, 07:06 AM

This is a tricky question.
$\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}$
Now set real part equal to real part and imaginary equal to imaginary.
$\frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4$
From that we get $p=\frac{1}{4}~\&~p=-1$ .

**drumist** · March 19th 2010, 08:25 AM

Originally Posted by Neverquit

I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.

There is also apparently another solution of 4, -1 which I can’t find.

How do I find it?

I think you are misinterpreting your own answer. When you solve the system of equations, you will find that

$p=1/4 ~~\mbox{ or }~~ p=-1$

These are two separate solutions, not a single solution. You need to find the value of $q$ that pairs with each of these solutions for $p$ . So, you need to plug each value of $p$ back into the system of equations and find the corresponding values of $q$ . Since $pq=1$ , it's a pretty straightforward calculation:

$p=1/4 \implies q=4$

$p=-1 \implies q=-1$

Therefore, the solutions are:

First solution: $p=1/4 \mbox{ and } q=4$

Second solution: $p=-1 \mbox{ and } q=-1$

You might also write this as:

$(p,q) = (1/4,4) \mbox{ or } (p,q) = (-1,-1)$

But you definitely would not say that the solutions are $(1/4,-1)$ and $(4,-1)$ .

**CaptainBlack** · March 20th 2010, 12:34 AM

Originally Posted by Neverquit

Hello,
I am having trouble with this question:

"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

Any help would be appreciated.

Regards,
Neverquit

$\frac{3+5i}{1+pi} =q+4i$

Multiply top and bottom of LHS by $1-pi$ to give:

$\frac{(3+5p)+(5-3p)i}{1+p^2}=q+4i$

Equate real and imaginary parts to get:

$3+5p=(1+p^2)q$

and:

$5-3p=4(1+p^2)$

Now the problem is to find all solutions to this pair of equations.

CB

**Deadstar** · March 20th 2010, 10:12 AM

Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?

Solving the equations I arrived at...

q - 4p = 3

pq = 1

gives

q=1/p,

So subbing that into the first equation will give you a polynomial (if you multiply both sides by p) which gives you p = 1/4 and -1.

Hence q = 4 and -1.

Solutions are (1/4,4) and (-1, -1)

Are these the only solutions? Why must you multiply the top and bottom by the conjugate?

**drumist** · March 20th 2010, 10:37 AM

Originally Posted by Deadstar

Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?

Your solution was correct. Multiplying a complex fraction by the conjugate of the denominator is probably just a habit for them. But it's not necessary at all in this particular problem.

**Neverquit** · March 20th 2010, 10:31 PM

Originally Posted by Plato

This is a tricky question.
$\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}$
Now set real part equal to real part and imaginary equal to imaginary.
$\frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4$
From that we get $p=\frac{1}{4}~\&~p=-1$ .

I think the solution that Plato gives using the congugate is what the text books author had in mind as the question is shortly after conjugates of complex numbers is explained.

Deastar, your solution still gives the same answer in the text book so it must be correct.

...........

**Pulock2009** · March 20th 2010, 10:36 PM

Originally Posted by Neverquit

Hello,
I am having trouble with this question:

"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i"

Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.

Any help would be appreciated.

Regards,
Neverquit

the basic idea behind questions of your type is equaslising real and imaginary parts.

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Complex numbers-finding real number pairs

found 1 solution but not the other

Using the conjugate

generalisation

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