last day (15 days later) » 

Mad Spaces
Mad Spaces
16:00
i am really impressed by your solution
it is very amazing and genius however the middle step is still unclear.
i am not that much of a pro
could you elaborate it please?
Astyx
Astyx
Did you see this ?
$(a+ib)(c+id) = ac -bd +i(ad+bc)$
Mad Spaces
Mad Spaces
Yes.. but i cant make the connection...
Astyx
Astyx
If you take the real part on both side, you find the real part of a product if the difference between the products of real and imaginary parts
You can find the same result without using complex numbers though, using the trigonometric identity $\cos (a+b) = \cos a\cos b - \sin a\sin b$
The path through complex numbers is just the way I remember it because I find it easier
Mad Spaces
Mad Spaces
@Astyx so ac is in this case our Re(first function * second)?
Astyx
Astyx
ac ?
Mad Spaces
Mad Spaces
16:06
in your formula above.
I am trying ot the mtch the symbols to the RE and IM
Astyx
Astyx
Ah
$a = Re(a+ib)$, $b = Im(a+ib)$, $c = Re(c+id)$, $d = Im(c+id)$
Mad Spaces
Mad Spaces
oh.
So in our case we wanted a * c (Re * Re
No wait..
user image
where did $ i(ad+bc)$ go?
Sorry if i am too dumb.
Astyx
Astyx
You're taking the real part, so the imaginary part goes away
Mad Spaces
Mad Spaces
oh!!!!!
Dude!!!
Amazing !!
Thanks ..! lol you are god
Astyx
Astyx
I didn't come up with this
I just learned it
Be sure to know how to do it without using complex numbers as well
In order to get familiar with it
Mad Spaces
Mad Spaces
16:17
Yes i know how to do this with trigenometrie as well, but this is really very nice solution.

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