Post by: tony3272 on November 10, 2011, 10:43:39 am
Post by: golden on November 10, 2011, 10:45:26 am
Would I lose all marks if I didn't do it that way but instead did:
Multiply top and bottom by the conjugate of the bottom, then do all the calculations etc. etc. and end up with invtan(answer)...
Would I get 1 mark at least?
Post by: chrysalis on November 10, 2011, 10:46:29 am
And yeah, 11pi/12.
Post by: HarveyD on November 10, 2011, 10:46:49 am
would I lose a mark? :S
Post by: kenny_ on November 10, 2011, 10:50:24 am
I put 11pi/12 + 2*pi*k
would I lose a mark? :S
You probably will cause Principle Arg -pi < x =< pi
Post by: Zebra on November 10, 2011, 10:52:10 am
Post by: dopplereffect on November 10, 2011, 10:58:52 am
Post by: Asx4Life on November 10, 2011, 10:59:59 am
Post by: b^3 on November 10, 2011, 11:12:07 am
Post by: tiktokcheekaboom on November 10, 2011, 11:14:10 am
Are there solutions up for this? I still can't think where I went wrong...
Post by: b^3 on November 10, 2011, 11:16:06 am
Post by: HarveyD on November 10, 2011, 11:18:35 am
i worked out
that the angle is -13pi/12
then below it wrote 11pi/12 (but didn't specify why)
then drew an arrow up to show that theta would be 11pi/12 + 2*pi*k for k element of real
since z = rcis theta
so I'd lose a mark for that last step yeah?
Post by: kenny_ on November 10, 2011, 11:18:49 am
Worked it out. When we did the angle for each part of the top and bottom oin polar form, for the top I took the angle as 2pi/3 when we should have used -pi/3 cause for the arg things tan is restricted to -pi/2 to pi/2.
arg is restricted from -pi to pi, 2pi/3 would still work, it's because the angle was in the 4th quadrant. positive x, negative y
Post by: tiktokcheekaboom on November 10, 2011, 11:19:38 am
DARN! D:
Post by: chrysalis on November 10, 2011, 11:20:59 am
Post by: Asx4Life on November 10, 2011, 11:21:53 am
Worked it out. When we did the angle for each part of the top and bottom oin polar form, for the top I took the angle as 2pi/3 when we should have used -pi/3 cause for the arg things tan is restricted to -pi/2 to pi/2.
I think its because the top part was 1-root3 i, and this is in the 4th quadrant, so it had to be -pi/3?
Post by: b^3 on November 10, 2011, 11:22:43 am
Worked it out. When we did the angle for each part of the top and bottom oin polar form, for the top I took the angle as 2pi/3 when we should have used -pi/3 cause for the arg things tan is restricted to -pi/2 to pi/2.
arg is restricted from -pi to pi, 2pi/3 would still work, it's because the angle was in the 4th quadrant. positive x, negative y
yep guys thats what happened then, dam I miss read that otehr negative :(
Post by: BoredSatan on November 10, 2011, 11:23:35 am
a bunch of the relatively smarter kids at my school had no idea how to do this question..
and i was sitting there wondering how this question was 3 marks..
Post by: BoredSatan on November 10, 2011, 11:24:15 am
2pi/3 wont work because thats in the 2nd quadrant.. you needed the 4th quadrant equivalent which was -pi/3Worked it out. When we did the angle for each part of the top and bottom oin polar form, for the top I took the angle as 2pi/3 when we should have used -pi/3 cause for the arg things tan is restricted to -pi/2 to pi/2.
arg is restricted from -pi to pi, 2pi/3 would still work, it's because the angle was in the 4th quadrant. positive x, negative yThats what I though but then why is the solutions not equal? I did minus the bottom from the top one.
yep guys thats what happened then, dam I miss read that otehr negative :(
Post by: naomizz on November 10, 2011, 11:30:07 am
hey guys, if i didn't convert the whole cartesian form into polar form, as in i didnt write down the modulus z value. i only found the argument for the numerator and denominator and find the arg of z by subtracting them off, would i get marks off?
Post by: b^3 on November 10, 2011, 11:31:36 am
Yeh I've realised now. I read the top as x negative and y positive. fml2pi/3 wont work because thats in the 2nd quadrant.. you needed the 4th quadrant equivalent which was -pi/3Worked it out. When we did the angle for each part of the top and bottom oin polar form, for the top I took the angle as 2pi/3 when we should have used -pi/3 cause for the arg things tan is restricted to -pi/2 to pi/2.
arg is restricted from -pi to pi, 2pi/3 would still work, it's because the angle was in the 4th quadrant. positive x, negative yThats what I though but then why is the solutions not equal? I did minus the bottom from the top one.
yep guys thats what happened then, dam I miss read that otehr negative :(
Post by: Sammmybear on November 10, 2011, 11:52:43 am
I converted the fraction to polar form, then simplified:
sqrt(2) cis(-(pi)/4)
2 cis (2(pi)/3)
= sqrt(2) cis(-11(pi)/12)
hence, Arg(z) = -11(pi)/12
What'd I do wrong??
Post by: kenny_ on November 10, 2011, 12:27:12 pm
For question 4 I got a different answer:
I converted the fraction to polar form, then simplified:
sqrt(2) cis(-(pi)/4)
2 cis (2(pi)/3)] supposed to be -pi/3, since it's 4th quadrant
= sqrt(2) cis(-11(pi)/12)
hence, Arg(z) = -11(pi)/12
What'd I do wrong??
Post by: ttn on November 10, 2011, 01:02:48 pm
why the hell did I put pi on the end for.
Do I still get one mark for getting that?
Post by: Sammmybear on November 10, 2011, 01:20:22 pm
For question 4 I got a different answer:
I converted the fraction to polar form, then simplified:
sqrt(2) cis(-(pi)/4)
2 cis (2(pi)/3)] supposed to be -pi/3, since it's 4th quadrant
= sqrt(2) cis(-11(pi)/12)
hence, Arg(z) = -11(pi)/12
What'd I do wrong??
I think my problem is that I said that 4+9=11 and it all went downhill from there. You don't NEED to convert it to -pi/3, you just end up with cis(-13pi/12) and then add 2pi to convert that to 11pi/12 as it has to be in Arg form.
Thanks anyway :)