My Specialist Maths Mega-thread :D

[monkeywantsabanana]

Hey guys, I'll have heaps and heaps of questions throughout the year so I made this mega thread so then I wouldn't have to make a new topic every time! 

My first question:

Show that

Is this how you're meant to approach it?




Seems like I'm going round in circles

[brightsky]

cos(-t) = cos(t), sin(-t) = -sin(t)

[Special At Specialist]

cis(-θ) = cos(-θ) + i*sin(-θ)
since cos(-θ) = cos(θ) and sin(-θ) = -sin(θ)
cis(-θ) = cos(θ) - i*sin(θ)
cis(-θ) = cos(θ) - sin(θ)*i

[monkeywantsabanana]

cis(-θ) = cos(-θ) + i*sin(-θ)
since cos(-θ) = cos(θ) and sin(-θ) = -sin(θ)
cis(-θ) = cos(θ) - i*sin(θ)
cis(-θ) = cos(θ) - sin(θ)*i

THANK YOU!

[monkeywantsabanana]

Come someone please explain why this is incorrect?

Express each of the following in the modulus–argument form,





I'm not too familiar with trig identities so the values found are obtained from the calc.

[nubs]

Come someone please explain why this is incorrect?

Express each of the following in the modulus–argument form,





I'm not too familiar with trig identities so the values found are obtained from the calc.

Your modulus is right, but you've taken your argument to be
The argument should be arctan()
Remember, the argument is the inverse tan of the imaginary component divided by the real component
So if z = a + bi
arg(z) = arctan(b/a)

[monkeywantsabanana]

Come someone please explain why this is incorrect?

Express each of the following in the modulus–argument form,





I'm not too familiar with trig identities so the values found are obtained from the calc.

Your modulus is right, but you've taken your argument to be
The argument should be arctan()
Remember, the argument is the inverse tan of the imaginary component divided by the real component
So if z = a + bi
arg(z) = arctan(b/a)

Wait, so what should I have done?



The calculator gives me

Where do I go from here? 

[brightsky]

first of all make sure you fully understand what a complex number is. from primary school through to year 11, we've always dealt with real numbers. these are numbers which can be located on a real number line (a horizontal line that goes from -infty to 0 to + infty). for complex numbers, we essentially add another 'dimension' - a vertical line - on top of the horizontal real number line. now the real number line has expanded to become something that resembles a Cartesian graph and we can plot numbers everywhere on the graph, e.g. (2,2), (3,-1) etc. the first 'x-coordinate' describes the real part of a complex number (Re(z)), and the 'y-coordinate' describes the imaginary part (Im(z)). recall that a complex number can be written in the form z = x+yi. this means that Re(z) = x and Im(z) = y, and we can plot the complex number as point (x,y) on the 'Cartesian graph' that i described above. (technical name for this graph is the Argand diagram).

now there is another way of expressing a complex number. the book calls it 'the polar form'. this form isn't concerned about the actual 'x and y coordinates' of the complex number z, but rather is concerned about the angle it makes with the x-axis and magnitude of the line connecting it and the origin. this form is generally written as z = r*cis(t), where r is the magnitude and t is the angle. suppose we already know a complex number in its cartesian form. hence we can already plot it on the Argand diagram at point (x,y). but how do we express it in polar form. first, we need to find r (the magnitude of the line connecting (x,y) and (0,0)). how do we do that? pythag! r = sqrt(x^2 + y^2). this magnitude is sometimes written as mod(z). now we need to find the angle t. how do we do that? trig! tan(t) = opp/adj = y-coordinate/x-coordinate = y/x. that means t = arctan(y/x).

now that that's out of the way, we refer back to your question. it is clear that in this case x = 1 and y = cot(t). (recall that cot(t) is just a fancy name for 1/tan(t)). now run through the procedure described above.

[monkeywantsabanana]

Please excuse my stupidity.

I have found r:

But, for theta, I don't understand why when i typed:

The calculator gives me

sorry...

[dc302]

@ brightsky, you know a lot for a yr 10

[Special At Specialist]

I was always taught to find the angle like so:
First find r where r = sqrt(x^2 + y^2)
cos(θ) = x/r where 0≤θ≤2pi (2 solutions)
sin(θ) = y/r where 0≤θ≤2pi (2 solutions)
Either find the solution that they both have in common and use that as your angle, or look at which quadrant your complex number is in and work it out from there.

But then I worked out:
tan(θ) = y/x where 0≤θ≤2pi (2 solutions)
Find the quadrant that the angle is in and that's your answer, without the need for calculating the modulus.

All the methods have their advantages.
In the first method, you can gain the intuition better and it is easier to transpose the equation and work backwards. In the second method, it is quicker and probably easier to solve.

[monkeywantsabanana]

Yeah, haha thanks, though, I do understand how to find the modus and the angle... I figured out the original question





So:


That's what I wanted.
Thanks all.

[monkeywantsabanana]

Find the cartesian equation for this:
 

[funkyducky]

Here's a clue for you: Let z=x+yi and square both sides to get rid of the modulus signs.

[monkeywantsabanana]

I did that but then i get up to:.... I'm suppose to sketch this too... if m is a random number, how am i supposed to? :|

Equating co-efficients: ,