Question

COMPLEX NUMBERS HELP NEEDED.
*question attached below* will give medal :)

Answer 1

I am having trouble with how to go about starting this problem since the numerators and denominators are either in Cartesian or complex numbers form. Any suggestions?

Answer 2

Have you heard multiplying by the conjugate?

Answer 3

Its like the additive inverse.

Answer 4

Yes

Answer 5

So should I multiply the first fraction by its conjugate?

Answer 6

Do that with the fraction with the complex number.

Answer 7

Alrighty!

Answer 8

Then, make them all have the same denominator by multiplying a fraction with another which gives the same denominator.

Answer 9

The one with the complex number in its denominator I should multiply by its conjugate?

Answer 10

Yes

Answer 11

That way you can get rid of the \(i\)

Answer 12

I'm a little confused as to how to make all of them have the same denominator by multiplying a fraction with another which gives the same denominator :S

Answer 13

Ok then. I will show the basic steps

Answer 14

Equation:
\[\frac{2y+4i}{2x+y}-\frac{y}{x-i}=0\]
Multiply \(\frac{y}{x-i}\) by the conjugate

Answer 15

Let me multiply it out :)

Answer 16

\[\frac{2y+4i}{2x+y}-\frac{y}{x-i}*\frac{x+i}{x+i}=0\]

Answer 17

Ok

Answer 18

This equals to:
\[\frac{2y+4i}{2x+y}-\frac{yx+yi}{x^2+1}=0\]

Answer 19

Okay, I got that when I multiplied :)

Answer 20

Good. Now, you need both to have the same denominator. Do you know how to do that?

Answer 21

Clueless :S

Answer 22

Thats fine ;)
Equation:
\[\frac{2y+4i}{2x+y}-\frac{yx+yi}{x^2+1}=0\]
Move the \(-\frac{yx+yi}{x^2+1}\) to the RHS
\[\frac{2y+4i}{2x+y}=\frac{yx+yi}{x^2+1}\]
Now use cross multiplication:
\[(2y+4i)(x^2+1)=(yx+yi)(2x+y)\]

Answer 23

Does that make sense?

Answer 24

Yes, it does :D

Answer 25

:D
Do that

Answer 26

Okay, give me a sec :)

Answer 27

\[2x^2y+4ix^2+2y+4i=2x^2y+xy^2+2ixy+iy^2\]

Answer 28

Boy, THATS complicated

Answer 29

Simplify equation: \[4ix^2+2y+4i−xy^2+2ixy−iy^22=0\]

Answer 30

I'm good up to that point :)

Answer 31

So. what do you need to do now?

Answer 32

Regroup and factorise?

Answer 33

Good. From there, you can create two equations to get both x and y

Answer 34

Let me work on it :)

Answer 35

Ok then

Answer 36

@dumbcow , no there aren't any posted answers :O

Answer 37

I'm stuck :/

Answer 38

We can input it directly to a calculator. Let me see what the numbers are given...

Answer 39

nevermind... you get a hyperbola for possible solutions

Answer 40

\[y=2(x-i)\]
According to \(\text{Wolfram Alpha}\)

Answer 41

\[x=\frac{1}{2}(y+2i)\]
According to \(\text{Wolfram Alpha}\)

Answer 42

The question said x and y are real values?

Answer 43

That's whats confusing me. I haven't studied \(\text{Hypebola}\) in detail yet, so I am not sure about the specific procedure.

Answer 44

@Ahsome , did you separate the real terms from "i" terms, then set each equal to 0
this is only way to make numerator 0
then from there, set the 2 equations equal --> real terms = imaginary terms

Answer 45

@dumbcow, No, I did not do that. But that sounds like the right thing to do.

Answer 46

From whiich step do I have to separate the real terms from the imaginary terms?

Answer 47

hold on sorry, this problem is giving me fits
as soon as i got the right equations, now im finding all these extraneous solutions
maybe im doing something wrong

Answer 48

Alrighty

Answer 49

ok i got it now finally
real terms:
\[2x^2 y +2y -2x^2 y - xy^2 = 0\]
imaginary terms:
\[4x^2 +4-2xy-y^2 = 0\]

you can simplify "real" equation
\[\rightarrow 2y - xy^2 = 0\]
\[\rightarrow 2 -xy = 0\]
\[\rightarrow y =\frac{2}{x}\]
sub that into "imaginary" equation
\[\rightarrow 4x^2 +4 -2x(\frac{2}{x}) - (\frac{2}{x})^2 = 0\]
\[\rightarrow 4x^2 - \frac{4}{x^2} = 0\]
\[\rightarrow 4x^4 = 4\]
\[\rightarrow x = \pm 1\]
There are 2 solutions:
(1,2)
(-1,-2)

Answer 50

wow! Thank you so much! It's all clear! :)

Answer 51

yw