The Student Room Group

arguments and radians (FP2)

Hi I'm in desperate need of help to do with the topic involving the modulus and argument in the further pure 2. I'm talking stuff like: tan(theta)= y/x in terms of radians. I keep looking at questions, methods, answers etc but to no prevail. It's weird because some questions I can do and others just seem to follow no pattern, i.e, with the method I've learnt. Help! Please!
Reply 1
If I can remember rightly to find the modulus of a complex number z=x+iyz=x+iy we calculate
Unparseable latex formula:

\sqrt{x^2+y^2



To work out the argument it is easiest to first draw the number on an Argand plane with a line connecting the point to the origin (think of a comlplex number as a co-ordinate (x,y)(x,y) with the real part as the x and the imaginary part as the y the argument is then the angle that the line makes with the positive x axis and so is always between -pi and pi. It always used to puzzle me why they defined the argument as being the angle the complex number makes with the positive x axis :confused: But anyway!

Drawing the complex number on the Argand plane will allow you to see the geometry of the situation and help you to find the angle required.

Also whenever you have to leave an answer in terms of an argument and say you have sine of an angle outside pi and -pi you have to keep adding 2pi to this angle until the new value is within -pi and pi. sin(θ+2kπ)=sinθ\sin{(\theta+2k\pi)}=\sin{\theta} and cos(θ+2kπ)=cosθ\cos{(\theta+2k\pi)}=\cos{\theta} for k an integer
Reply 2
What's an example question that you're stuck on? If we help you with one thing you're stuck on, then hopefully that'll help you with other similar things.
Reply 3
Yes that would be useful. An example is worth a thousand theorems and definitions :smile:
Reply 4
Pascal
Hi I'm in desperate need of help to do with the topic involving the modulus and argument in the further pure 2. I'm talking stuff like: tan(theta)= y/x in terms of radians. I keep looking at questions, methods, answers etc but to no prevail. It's weird because some questions I can do and others just seem to follow no pattern, i.e, with the method I've learnt. Help! Please!


This website may help:

http://www.geocities.com/maths9233/Complex/Complex.html
Reply 5
The best way I found for my FP1 exam (we're obviously doing different boards) was to do the following.

Firstly, plot the thing on an Argand diagram

then:

Complex no. : a+bj
Let alpha = atan (|b/a|) -- in radians (Check calculator mode, and make sure it's the magnitude of b/a, not the numerical value!!)

The argument of the complex number depends on which quadrant it is in:

First quad: arg = alpha
Second quad: arg = pi-alpha
Third quad: arg = alpha-pi
Fourth quadrant: arg = -alpha

It's intuitively obvious once you've drawn the argand diagram. Anyway, I hope that helps!
Reply 6
darth_vader05
hey henry :biggrin: (long time no speak!)

i think its important to draw a diagram first. my teacher drilled it into me, and you can check the signs

Hey!! Gosh, it has been long! I think I might go and waste some time on HP thread this Summer, given that I've got nothing better to do. And deffo agree with drawing the diagram :biggrin:
Reply 7
Erm... Lengthy socialising?? :biggrin:
Reply 8
darth_vader05
and to the OP i think the range for the angle should be from pi to -pi

Which is encorporated into my method :p:
Reply 9
henryt
The best way I found for my FP1 exam (we're obviously doing different boards) was to do the following.

Firstly, plot the thing on an Argand diagram

then:

Complex no. : a+bj
Let alpha = atan (|b/a|) -- in radians (Check calculator mode, and make sure it's the magnitude of b/a, not the numerical value!!)

The argument of the complex number depends on which quadrant it is in:

First quad: arg = alpha
Second quad: arg = pi-alpha
Third quad: arg = alpha-pi
Fourth quadrant: arg = -alpha

It's intuitively obvious once you've drawn the argand diagram. Anyway, I hope that helps!


Yup, that's the best method there is! :biggrin: Definitely draw a diagram, otherwise its really hard to do.
Reply 10
Ok fair enough so I draw teh Argand diagram and I work out for instance that tan(theta)= 90 degrees approx from looking at the diagram. How do I now convert this to a nice looking radian, e.g something like sqrt(pi). I can't grasp how to change the degrees to radians and get the radian to look orderly or neat in any way. Definately drawing the Argand diagram helps but I'm still in a pickle!
degree to radans is x/180*pi, since pi= 180 degrees. But like Arvi said there are certain ones you need to know from the top of your head (30, 45, 60, 90, 180 etc periodically). Learn the triangles.

Also note, the argument has to be greater than -pi, it cant be -pi itself.
Reply 12
i think i understand them a bit more now i'm going to go tackle some questions. Thanks everyone and i'll post a thread if i get stuff, which i invariably will!
Reply 13
henryt
The best way I found for my FP1 exam (we're obviously doing different boards) was to do the following.

Firstly, plot the thing on an Argand diagram

then:

Complex no. : a+bj
Let alpha = atan (|b/a|) -- in radians (Check calculator mode, and make sure it's the magnitude of b/a, not the numerical value!!)

The argument of the complex number depends on which quadrant it is in:

First quad: arg = alpha
Second quad: arg = pi-alpha
Third quad: arg = alpha-pi
Fourth quadrant: arg = -alpha

It's intuitively obvious once you've drawn the argand diagram. Anyway, I hope that helps!


Ok what's alpha in terms of this?
Reply 14
I'm not sure I understood your question... I defined alpha = atan(|b/a|), so it's just an angle. You then need to adjust the angle alpha depending on the quadrant (explained in Prev. Post) to get the argument.
Reply 15
Ah, so it's just theta or x or whatever then. Ok, ok. I don't know if this is just me not getting this but how the hell do you change the angle into radians. I know theat you divide by 180 and multiply by pi but then I just end up with really stupid numbers. How would I work out for example that the argument of -3+i is sqrt of pi/2. Which is the real answer (it's out the text book). How do you get nice, neat looking radians such as this. If I do it the way I think I should, y/x which is -1/3 and I turn that into radians I don't get sqrt of pi/2.