mod arg form

z1 - z2 = cos(x1 -x2) + isin(x1-x2)

This is not true. Where is this coming from?

Anyway, there is a typo on the second line. Obviously it should read

$\dfrac{(\cos 2x + i\sin 2x)(\cos 9x + i\sin 9x)}{(\cos 9x - i \sin 9x)(\cos 9x + i\sin 9x)}$

but if they did it like that, that following work wouldn't be the same. have a look at the answer again.

@RDKGames is right. Its a typo. You're multipying by the denom conjugate on the top and bottom. The bottom is unity magnitude (as its a unit vector). The top becomes
cos(11x) + isin(11x)
by addition formulae or more simply by de Moivre (if you've done it).

look at their answer - they've not multiplied through using the conjugate

look at their answer - they've not multiplied through using the conjugate

look at their answer - they've not multiplied through using the conjugate

the textbook

Yes they have!

cos(9x) + isin(9x) is the conjugate of cos(9x) - isin(9x)

why is it done this way? I know cos(x) = cos(-x) and -sin(x) = sin(-x)

and that z1/z2 = z1 - z2 = cos(x1 -x2) + isin(x1-x2) = cos(11x) = isin(11x) - the same answer as solution bank.

they've not even done: (z1/z2)(z*2/z*2)

Yes what they show is the result for division, but you said z1 - z2 instead of z1/z2 therefore I said your claim is incorrect.

They have, its a typo.
Line 3 is a scalar (magnitude - squared) on the denom. They've multiplied by the conjugate.

We are talking about Q17?

It's a typo. Just multiply top and bottom by the complex conjugate of the bottom.

expand out and it works. where do you think cos^2+sin^2 has come from?

I said z1/z2 = z1 - z2

Yeah, hence automatically saying that division is the same as subtraction is incorrect.

Anyway, hopefully you understand how having z1 - z2 is incorrect in that line.

yeah, I think I do

You think?

Well ok, let me put it this way;

It's like me saying $\dfrac{10}{2} = 10 - 2$

It's just a simple nonsensical error, nothing more to it. Just be careful in the future with what you write, that's the take-home message.

sorry, I was confused with arg(z1/z2) = arg(z1) - arg(z2)