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# Fp1 watch

1. (Original post by IrrationalRoot)
Fixed, Latex is being stupid for some reason.
You wrote ) instead of } in your first set of brackets, which is why it threw the error.
2. (Original post by morgan8002)
You wrote ) instead of } in your first set of brackets, which is why it threw the error.
Ahhh, thanks, my eyesight is really bad.
3. (Original post by Zacken)

If you draw this sort of diagram it becomes clear. The red angle you find using trigonometry, and then the green angle is your argument which you find by doing pi-red angle.
Yep this is what I do. No point working out inverse tan every time, just confuses a lot of students.
4. (Original post by IrrationalRoot)
Yep this is what I do. No point working out inverse tan every time, just confuses a lot of students.
I do this every time too! It's really easy to just jot a tiny triangle in the corner of your page as well.
5. (Original post by Zacken)

If you draw this sort of diagram it becomes clear. The red angle you find using trigonometry, and then the green angle is your argument which you find by doing pi-red angle.
Ahh thats much clear! Thanks Zack
6. (Original post by Ayaz789)
Ahh thats much clear! Thanks Zack
Cheers.
7. (Original post by Zacken)
Cheers.
So what if it was -1-3i , itd be in the 4th quadrant which would be 2 Pi - theta & theta would be tan-1(3/1)
8. (Original post by Ayaz789)
So what if it was -1-3i , itd be in the 4th quadrant which would be 2 Pi - theta & theta would be tan-1(3/1)
You need to sketch(!) - if you sketched, you'd clearly see that it's in the third quadrant and hence the argument would be pi + theta where theta = tan-1 (3/1)
9. (Original post by Zacken)
You need to sketch(!) - if you sketched, you'd clearly see that it's in the third quadrant and hence the argument would be pi + theta where theta = tan-1 (3/1)
Like 4th quadrant is like the bottom left imo haha & okay i get you
10. (Original post by Zacken)
You need to sketch(!) - if you sketched, you'd clearly see that it's in the third quadrant and hence the argument would be pi + theta where theta = tan-1 (3/1)
When you always do the inverse of tan with complex numbers is it always going to be with positve numbers?
11. (Original post by Ayaz789)
Like 4th quadrant is like the bottom left imo haha & okay i get you
Oh, okay.

(Original post by Ayaz789)
When you always do the inverse of tan with complex numbers is it always going to be with positve numbers?
With this method, yeah.
12. (Original post by Zacken)
Oh, okay.

With this method, yeah.
Ahh okay thanks so say if it was 1-3i , itd be bottom right quadrant & it'd be theta= tan(3/1) therefore argument would equal 2Pi - ANSWER?
13. (Original post by Ayaz789)
Ahh okay thanks so say if it was 1-3i , itd be bottom right quadrant & it'd be theta= tan(3/1) therefore argument would equal 2Pi - ANSWER?
Yep.
14. (Original post by Zacken)
Yep.
Great!
15. (Original post by Ayaz789)
Great!
Cheers.

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