Complex Number Problem Help

[Pruko]

I’m trying to set the real = real, and imaginary = imaginary; but I can’t manipulate the constants and variables correctly to set-up the correct equations for x and y, please help. Thank you in advance.

 

[Deleted member 4993]

I’m trying to set the real = real, and imaginary = imaginary; but I can’t manipulate the constants and variables correctly to set-up the correct equations for x and y, please help. Thank you in advance.

You said:

".....but I can’t manipulate the constants and variables correctly......"

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[pka]

Have a look at this.

 

[Steven G]

Multiply both sides by i-2.

 

[Pruko]

Multiply both sides by i-2.

Why not by (-2 - i)? Or, is it the same?

 

[pka]

Why not by (-2 - i)? Or, is it the same?

I think that is a typo. Surely its should be \(-2-i\).
Look at reply # 3

 

[skeeter]

[MATH]-2-i = -(2+i)[/MATH]
[MATH](3-2x)+(-1+2y)i = -(2+i)(2-i)[/MATH]
expand the RHS, then equate the real and imaginary components

 

[Steven G]

I think that is a typo. Surely its should be \(-2-i\).
Look at reply # 3

I stick by my statement of i-2.
-2-i = -(2+i) and a conjugate is -(2-i) = i-2

 

[Steven G]

Why not by (-2 - i)? Or, is it the same?

Are they the same? Clearly -2-i has -i while i -2 has i. i and -i are not the same.

 

[HallsofIvy]

Notice that (-i- 2)(2- i)= -(2+ i)(2- i) and now remember that \(\displaystyle (a+ b)(a- b)= a^2- b^2\).