Principle Argument:

[Asx4Life]

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?

[b^3]

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

Thats 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

[BoredSatan]

im surprised at how many people dont know how to convert from cartesian to polar

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..

[BoredSatan]

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

Thats 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

2pi/3 wont work because thats in the 2nd quadrant.. you needed the 4th quadrant equivalent which was -pi/3

[naomizz]

dang! was staring at this question for a long time in the exam, but i guess i'll still get a mark off. cause i wrote 11pi/12 but then i concluded that principle argument is 11pi/12 + 2kpi at the end! completely forgot that principle argument is the restricted angle! FML!
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?

[b^3]

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

Thats 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

2pi/3 wont work because thats in the 2nd quadrant.. you needed the 4th quadrant equivalent which was -pi/3

Yeh I've realised now. I read the top as x negative and y positive. fml

[Sammmybear]

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)

= sqrt(2) cis(-11(pi)/12)

hence, Arg(z) = -11(pi)/12

What'd I do wrong??

[kenny_]

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??

[ttn]

shit, I left it as arctan(root3 - 2) * pi ....
why the hell did I put pi on the end for.
Do I still get one mark for getting that?

[Sammmybear]

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