Complex Division: Solving Proof for Dividing 2 Complex Numbers

[Monochrome]

I'm working on the proof that two complex numbers can be divided from Alhford and I'm completely s(t)uck.

I've gotten as far as:

a = gx - dy
b = dx +gy

from
(a+ib) / (g + id)

where

a+ib = (g + id)(x+iy)

I've managed to get

$$x={\frac {b-{\it gy}}{d}}$$

and done the same for y (latex is a bit hard for me right now so I can't really show both and I'd like to have a go at it myself when I get what is done for x)

But the book says I'm mean to get:
$$x={\frac {{\it ag}+{\it bd}}{{g}^{2}+{d}^{2}}}$$
Help?

 

[AKG]

I've gotten as far as:

a = gx - dy
b = dx +gy

from
(a+ib) / (g + id)

where

a+ib = (g + id)(x+iy)

I've managed to get

$$x={\frac {b-{\it gy}}{d}}$$

and done the same for y (latex is a bit hard for me right now so I can't really show both and I'd like to have a go at it myself when I get what is done for x)

But the book says I'm mean to get:
$$x={\frac {{\it ag}+{\it bd}}{{g}^{2}+{d}^{2}}}$$
Help?

You want to solve for x and y. If you have an expression for x that involves y, you haven't really solved for x. You have deduced two equations

a = gx - dy
b = dx +gy

You used the second one to isolate x and got

x = (b-gy)/d

Now plug this into the first equation, solve for y. Then sub your solution for y back into your expression for x, and solve for x. Note, how do you know you're allowed to divide by d, as in x = (b-gy)/d? What if d = 0?

 

[Monochrome]

Oh bloody hell, I had that at the start but didn't see the division by d continued to g. Thanks.

And about d = 0, the book says we're to assume that d isn't zero, than it just becomes normal division.