Complex Numbers.

Raskolnikov

Simple question (well I think so at least!)

x+(x+y)i = 3+8i

The way I would have done it was . . .

[PHP]
x + (x+y)i = 3 + 8i
(x + (x+y)i) - (3 + 8i) = 0
(x - 3) + ((x+y)i - 8i) = 0
x=3 (x+y)-8 = 0
(3+y)-8 = 0
(-24-8y) = 0
-24 = 8y
-3 = y[/PHP]

Apparently the answer is x=3 and y=-2 so I'm doing something wrong. I haven't done complex numbers for two years so be kind and point out my mistake

David19

x+(x+y)i = 3+8i

Therefore x = 3 and x+y = 8

Therefore y = 5.

Y can't be -2. If you put -2 in for y you get 3+1i on the left side.

David19

Btw the problem with your way is here:

You have (3+y)-8 = 0

Then you get (-24-8y) = 0
This is wrong. its not 8 multiplied by (3+y). Its (3+y)-8 which equals (3+y-8) which equals (y-5).

You see?

Raskolnikov

David19 wrote:

Btw the problem with your way is here:

You have (3+y)-8 = 0

Then you get (-24-8y) = 0
This is wrong. its not 8 multiplied by (3+y). Its (3+y)-8 which equals (3+y-8) which equals (y-5).

You see?

:eek:

I see where you're coming from.

I'm informed though, that the answer is y = -2, I actually had y = 5 to begin with but I thought that was all wrong.

Can anyone give closure that the answer is indeed y = 5?

D-Generate

Yeah y = 5.

Text books often have the wrong answers so they are not taken as gospel.

ecksor

Can we assume that x and y are real?

M@lice

its y=5 for me too. Given x and y are real of course. But if x and y could be complex i'm pretty sure there would be infinitely many answers probably with some linear correspondance between x and y.