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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.
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.
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.
Reply 23
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Zacken
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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.
Reply 24
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Lychee628
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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
Reply 25
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Zacken
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Original post by Ayaz789
Ahh thats much clear! Thanks Zack


Cheers.
Reply 26
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Lychee628
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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)
Reply 27
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Zacken
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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)
Reply 28
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Lychee628
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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
Reply 29
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Lychee628
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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?
Reply 30
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Zacken
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Original post by Ayaz789
Like 4th quadrant is like the bottom left imo haha & okay i get you


Oh, okay. :tongue:

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.
Reply 31
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Lychee628
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Original post by Zacken
Oh, okay. :tongue:



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?
Reply 32
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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.
Reply 33
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Lychee628
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Original post by Zacken
Yep.

Great!
Reply 34
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Zacken
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Original post by Ayaz789
Great!


Cheers.

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