Finding X, Y in complex numbers.

[TheKittehCP]

Homework Statement

8i = ( 2x + i ) (2y + i ) + 1

The final answers is [x =0, 4]
[y=4, 0]

Homework Equations

The Attempt at a Solution

The final answer in the book is stated as above but if I follow the solution I will get the real parts which would
Equal 0=4xy meaning that both x and y when solving for either wil equal zero. If I continue the result I will get is

[X= 0, Y= 4]

I get one solution instead of 2 as provided in the book..

 

[rude man]

x=0 and y=4 is indeed one solution. There is another one and it's very similar to the first. Hint: note the symmetry of the real and imaginary equations in terms of x and y!

 

[abitslow]

I don't see your attempt at a solution...?
If you are given number = a *b +another number, then it is obvious that number - another number = a*b
In the case of complex numbers, if we call the Real part of x, Re(x), and Im(x) for the imaginary part so that x = Re(x)+Im(x), then it is straightforward that:
Re(number-another number) = Re((2x+i)(2y+i)) and
Im(number-another number) = Im((2x+i)(2y+i)) and
this gives you two equations and two unknowns.

 

[rude man]

OK, how did you arrive at x=0, y=4? (Your post 1). Show every step.

 

[rude man]

I don't see your attempt at a solution...?
If you are given number = a *b +another number, then it is obvious that number - another number = a*b
In the case of complex numbers, if we call the Real part of x, Re(x), and Im(x) for the imaginary part so that x = Re(x)+Im(x), then it is straightforward that:
Re(number-another number) = Re((2x+i)(2y+i)) and
Im(number-another number) = Im((2x+i)(2y+i)) and
this gives you two equations and two unknowns.

x and y are real numbers.

 

[TheKittehCP]

OK, how did you arrive at x=0, y=4? (Your post 1). Show every step.

My answers X=0 and Y=4 were correct, I talked to the teacher and he said the final answers given were wrong

Here's how I came up with it:

8i = 4xy + 2xi + 2yi + i^2 + 1

0 + 8i = 4xy + 2xi + 2yi -1 + 1 [ Since i^2 equals -1 ]

Real part = real part thus

0=4xy

X = 0/4y

X=0

Imaginary part = imaginary part thus

8 =2x + 2y

-2y = 2x -8

2y = 8-2x

Y = (8-2x)/2.. Now substitute with X=0

Y = (8-0)/2

Y= 4

[ X= 0, Y=4]


This is the correct solution. Thanks for the help everyone ^^

 

[LCKurtz]

My answers X=0 and Y=4 were correct, I talked to the teacher and he said the final answers given were wrong

Here's how I came up with it:

8i = 4xy + 2xi + 2yi + i^2 + 1

0 + 8i = 4xy + 2xi + 2yi -1 + 1 [ Since i^2 equals -1 ]

Real part = real part thus

0=4xy

X = 0/4y

X=0

$4xy=0$ means either $x=0$ or $y=0$. You only have one of the two answers.

Imaginary part = imaginary part thus

8 =2x + 2y

-2y = 2x -8

2y = 8-2x

Y = (8-2x)/2.. Now substitute with X=0

Y = (8-0)/2

Y= 4

[ X= 0, Y=4]


This is the correct solution. Thanks for the help everyone ^^

It is an incomplete solution. $x=4,~y=0$ also works as suggested in post #2.

 

[rude man]

Looks like LC did the rest for you. It is correct. If your teacher said the two answers are wrong then HE is wrong.

 

[TheKittehCP]

I also did that when I didn't get the answer same as mine but I think the difference is that X doesn't equal 0 and 4 in the same formation; X either equals 0 or 4 depends on what you solve for.. It doesn't equal 0 and 4 when we solve for the same variant such as

x (x-4)=0

Where X will equal both 0 and 4 and get final answer as [x = 0, 4] instead of [x =0] only or [x=4] only.. I talked to the teacher about the possibility of when X equals 0
When we solve for x at 4xy = 0 and when it equals 4 when we solve for Y at the same part and he said they're both correct.8