Comments: 
From: (Anonymous) 20070415 01:48 pm (UTC)
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Hi, I'm not sure how to use this site really, but I was after some maths help?! Have looked everywhere on the net and can't seem to find a website...
Would you be able to show me the working out in how to solve this problem? (Equality of Complex numbers)
2x + 3yi = x 6i
Please & Thankyou !!!!
Hi. Complex numbers is not in the realm of my current knowledge, as you may know if you are aware of the general syllabus content of maths GCSEs, which as explained in my post I'm currently studying for. I have however posted your question on UMMF, the forum from which this LJ spin off was created, where there are many people much smarter and more knowledgable than me. If you don't have a UMMF account, you won't be able to see responses as they occur, but I'll keep you posted. :) I apologise for the lateness of my response, I switched off comment emails a while ago, as they were sending repeats emails for the same comment, thus clogging my inbox, and thus I wasn't aware of it until now. Thanks for visiting ummf_on_lj, by the way. If you do become an LJ user, please feel free to join the community!
The following is from the UMMF user snowyowl and is so far not been shown to be incorrect:
2x + 3yi = x 6i 3x+3yi=6i 3(x+yi)=6i x+yi=2i x+(y+2)i=0 x=(y+2)i So I think that it is true when x=(y+2)i
Hopefully this helps. :)
More from snowyowl:
As y=2+xi 1) If x and y is real, As y is real, Im(y)=0 therefore xi=0 But x is real, therefore x=0, then we get y=2 2) If x is complex (not real), y is real As same as 1) xi=0 Let x=a+bi, a,b is real. xi=ai+bi(i)=aib But xi is real, therefore Im(xi)=Im(aib)=a=0 Therefore a=0, b is real From a=0, b is real x=bi, b is any real number, which makes y=b2 3) If x is real, y is complex (not real) From y=2+xi If x=0, from (1) y=2 If x is not 0, then, Im(y)=Im(2+xi)=x which is not 0 Re(y)=Re(2+xi)=2 which is not 0 therefore y is complex x is any real, y=2+xi 4) If x and y is complex (not real) Let x=a+bi y=c+di when, a,b,c,d are real From y=2+xi c+di=2+(a+bi)i c+di=2+aib c+di=(b2)+ai therefore c=b2, d=a So y=c+di=(b+2)+ai
So the answers are when 1) x=0, y=2 2) x=ki, y=k2 when k is any real number 3) x=k, y=2+ki when k is any real number 4) x=a+bi, y=(b+2)+ai when a,b are any real numbers
But from 4) If a=b=0 x=0, y=2 which is the answer from (1) If a=0 x=bi, y=(b+2)=b2 which is the answer from (2) If b=0 x=a, y=2+ai which is the answer from (3)
Therefore the answers of the equation 2x+3yi=x6i are when x=a+bi, y=(b+2)+ai; a,b are any real numbers.
Hope this makes sense!
From: (Anonymous) 20070623 12:22 pm (UTC)
Thankyou Heaps! :)  (Link)

Thank you so much for the help!!! :) GREATLY appreciated...no probs about the late reply. :)
 From: yayworthy 20070623 12:28 pm (UTC)
Re: Thankyou Heaps! :)  (Link)

No problem! =]  