complex no. q (1 Viewer)

[AGB]

can someone please explain this to me??

the following is from a ext 2 book....

z(3i-1) = 5(-3+2i), divide by (3i-1), change the signs.

therefore z=
5(3-2i)
1-3i

i dont understand why you change the signs in the numerator and denominator when you divide by (3i-1)

EDIT: obviously there is more to the prob, but i just dont get this step...

 

[abdooooo!!!]

what sign?

if you change the sign for both numerator and denominator then you are not changing signs at all if you know what i mean, ie two negative makes positive...

its just a different way of expressing a number since complex numbers are normally written in the form of a + ib, ie real number first and imaginary second.

then you proceed in the question by rationalisation the denominator, ie timing it by its conjugate and leaving a real number and a imaginary number in the form of a + ib or x + iy.

or is there more to it? anyone know?

 

[AGB]

yeh i get that.....but in the book it specifically says "change the signs"

it works either way so i guess you're right then abdooooo

 

[+Po1ntDeXt3r+]

i think its just to reduce errors.. and looks neater..
=P

 

[Rorix]

Well, changing the signs makes the real component positive, but I doubt you'd lose any marks for that sort of thing.

 

[Xayma]

I dont see any specific reason unless it is for neatness (ie so you dont have to times by (-1-3i)/(-1-3i) instead of (1+3i)/(1+3i) )

 

[Rorix]

well, 5(-3+2i)/(-1+3i) is arguably not as neat as the answer

 

[KeypadSDM]

Originally posted by AGB
EDIT: obviously there is more to the prob, but i just dont get this step...

Maybe you change the sign to make the rest of the problem simpler?