Problem in Complex Number Question (1 Viewer)

[Mill]

Martin, look at this thing made by Avicenna:

http://www.boredofstudies.org/community/attachment.php?s=&postid=487446

Or ask a tutor to explain.

 

[martin310015]

ok i got it thanx

 

[Avicenna]

Attention: Mill has pointed out that i made an error in my solution...

After working out tan^-1(a/4b)+tan^-1(4b/a) to be pi/2 i forgot to substitue it back into cis[tan^-1(a/4b)+tan^-1(4b/a)]. So :

4/3xcis[tan^-1(a/4b)+tan^-1(4b/a)]
=4/3xcis[pi/2]
=4/3x[cos(pi/2)+isin(pi/2)]
=(4/3)i

 

[KeypadSDM]

Originally posted by ae
(4a+16ib) / (12b-3ia)

= 4(a+4ib)/ -3i(a+4ib)

= -4/3i * i/i

= 4i/3

When the solution's a 3 liner, why bother wasting a page of writing?

 

[Mill]

You are required to do it 2 ways.

The method you quoted is short Keypad.

The other 3 solutions, however, are all pretty much equally long.

 

[Grey Council]

out of curiosity, IS it possible to do via ordered pairs? I remember doing that chapter on ordered pairs, and there was division in it. AND it was quite simple.

humph, can't remember how to do it off the top of my head.

btw, ordered numbers are still in the syllabus.

 

[:: ck ::]

ordered pairs 0.o