Question

Define \( \oplus \text{ on } \mathbb{R} X \mathbb{R} \) by setting \( (a,b) \oplus (c,d)=(ac-bd, ad+bc) \)
Show that the function h from system \( (\mathbb{C}, \cdot) \to\ (\mathbb{R} x \mathbb{R}, \oplus) \) given by h(a+bi)=(a,b) is a one to one function from the set of complex numbers that is onto \(\mathbb{R} x \mathbb{R}\) and is operation perserving

Answer 1

I have no background in complex numbers whatsoever.

Answer 2

then this is going to be a treat

Answer 3

but as i wrote before this is how you construct the complex numbers out of the real numbers
the map you want is a simple one, takes \(a+bi\to (a,b)\)

Answer 4

right u started explaining it b4 but i so didnt follow cuz i never really studied it

Answer 5

it is certainly onto , and there is not much to show here
what you have to do is pick an element of \(\mathbb{R}\times \mathbb{R}\) and show that it comes from some complex number,
but it is so simple as to almost be confusing
if \((a,b)\in \mathbb{R}\times \mathbb{R}\) then it comes from the complex number \(a+bi\) that is \(h(a+bi)=(a,b)\) and so it is onto

Answer 6

so every complex number comes from a real number meaning
\( \mathbb{C} \subseteq \mathbb{R} \) ?

Answer 7

oh no this is not \(\mathbb{R}\)

Answer 8

it is \(\mathbb{R}\times \mathbb{R}\)

Answer 9

ohhh right

Answer 10

it is just a way of constructing them

Answer 11

one to one is easy too, because \(a+bi=c+di\iff a=c, b=d\)

Answer 12

that is, two complex numbers are equal iff their real parts and imaginary parts are equal

Answer 13

right got that

Answer 14

so if \(h(a+bi)=h(c+di)\) that means \((a,b)=(c,d)\) which in turn means \(a=c\) and \(b=d\) and so \(a+bi=c+di\)

Answer 15

last thing you have to show is
\[h((a+bi)(c+di))=h(a+bi)\oplus(c+di)\]

Answer 16

where the multiplication on the left is complex number multiplication, and the multiplication on the right is the \(\oplus\) multiplication defined earlier
it work for sure

Answer 17

Thanks @satellite73

Answer 18

yw

Answer 19

you do know how to multiply two complex numbers together right? that is all that is really needed here

Answer 20

hahah nooo but ill research that

Answer 21

really?

Answer 22

I have never touched complex numbers in my life

Answer 23

idk in HS we never covered it and this is the first time in university that it is coming up

Answer 24

since \(i=\sqrt{-1}\) you get
\[(a+bi)(c+di)=(ac-bd)+(ac+bd)i\] pretty much identical to what you have on the right, with the plus sign in the middle replaced by a comma

Answer 25

ok that was a typo
you get
\[(a+bi)(c+di)=(ac-bd)+(ad+bc)i\]

Answer 26

(a+bi)(c+di)=(ac−bd)+(ad+bc)i
y is the i only after the second term?

Answer 27

ohhhh i get it u just grouped it like that

Answer 28

Thanks satellite for having patience with me

Answer 29

didnt go to a good hs so i am missing the basics in math whtvr

Answer 30

you ca multiply out using the distributive law as well and you get the term \(bdi^2\) but since \(i^2=-1\) it sure in to \(-bd\)

Answer 31

gotcha Thanks :DDDDDDDDDDDDDDDDD

Answer 32

yw