Challenge complex numbers problem for current students (2 Viewers)

[CM_Tutor]

Suppose that z is a complex number with modulus 1 and argument @ (ie z = cos@ + isin@).

(a) Show that z^n - 1 = 2i * sin(n@ / 2) * [cos(n@ /2) + isin(n@ / 2)], for n a positive integer.

(b) Hence, or otherwise, show that
z + z^2 + z^3 + ... + z^n = {sin(n@ / 2) / sin(@ / 2)] * [cos[(n + 1)@ / 2] + isin[(n + 1)@ / 2]}

(c) Hence, find an expression for cos@ + cos(2@) + cos(3@) + ... cos(n@)

(d) Prove that |cos@ + cos(2@) + cos(3@) + ... cos(n@)| <= |cosec(@ / 2)|

 

[CM_Tutor]

(e) Solve the equation cos@ + cos(2@) + cos(3@) + cos(4@) = 0 over the domain 0 <= @ <= 2 * pi

 

[spice girl]

can u post challenge problems for ex-students too?

i'm starting to forget my 4u alreadi

 

[CM_Tutor]

If you want to have a go, by all means do so. All I meant was that I didn't want past students to jump in and give the solutions - rather, I wanted the current students to have a chance to have a go and try and solve them themselves.

PS: There are some bits of 4u that you will never use in Uni maths - like conics - and others - like complex numbers, integration and polynomials - that will keep popping up.

PPS: Why are you so keen on the Cr - Cr quadruple bond, and have you been studying chromium acetate recently?

 

[spice girl]

hehe nono not this problem. i meant harder problems.

and besides being a med student i'm deprived of even uni maths. how boring

 

[CM_Tutor]

Was there any particular topic you wanted to think about? In the meantime, here's a couple of things for you to think about:

1. We know that x^0 = 1, and 0^x = 0, so what is 0^0?

2. Bill and Ted play a game with a fair coin, where each toss the coin n times. Show that the probability that they get the same number of heads is (2n)! / [2^(2n) * (n!)^2]

 

[:: ck ::]

cisn@ -1 = 2isin(n@/2)cis(n@/2)
square both sides

cis2n@ - 2cisn@ + 1 = -4sin²(n@/2)cis(2n@)
= -4cis(2n@)( (1-cosn@)/2 )
= -2cisn@ + 2cisn@cosn@
= 2cosn@(cisn@) - 2cis(n@)
= 2cosn@(cosn@ + isin n@) - 2cis(n@)
= 2cos²n@ + i(2cos n@ sin n@) - 2cis(n@)
= cos2n@+1 + isin2n@ - 2cisn@
= cis2n@ -2cisn@ + 1

LHS^2 = RHS^2

 

[CM_Tutor]

You have proven that LHS^2 = RHS^2. That isn't the same as proving LHS = RHS. ie. Your answer is incomplete - if marked properly, it would not get full marks.

PS: As a piece of advice to all, be very careful in using cis, or better yet, don't use it at all - its amazing how many people stuff up questions by using this, especially in working like that above where there are cis's and cos's. (Ryan, I'm not saying your working is wrong, just offering a comment that you (and everyone else) are free to take or leave as they choose.)

 

[:: ck ::]

well if i prove LHS^2 = RHS^2

can i just say square root both sides?

 

[Mill]

LHS = - RHS

 

[CM_Tutor]

Originally posted by :: ryan.cck ::
well if i prove LHS^2 = RHS^2

can i just say square root both sides?

No, as Mill points out, you need to address the possibility that LHS = - RHS after you square root both sides - I'm not saying that determining whether LHS is RHS or - RHS is difficult (it isn't), just that your answer is incomplete until you do so.

As an illustration, suppose you were asked to prove that x = sqrt(2) satisfied an equation. An answer showing x^2 = 2 would be incomplete, as it would show that x = sqrt(2) or -sqrt(2), and would then need to establish either that x = -sqrt(2) is wrong, or that both are true.

 

[:: ck ::]

oh yeh silly me... squarin both is stupid >.<

newaiz ive doen it now

(a) Show that z^n - 1 = 2i * sin(n@ / 2) * [cos(n@ /2) + isin(n@ / 2)], for n a positive integer.

cisn@ - 1 = 2isin(n@/2)(cos(n@/2) - 2sin²(n@/2)
= isin(n@) + cosn@ - 1
= cisn@ -1

LHS = RHS

 

[Grey Council]

well done Ryan. hheeh, imagine how hard the other ones are. >.<

 

[:: ck ::]

lol thx it wasn that hard i just made it hard on my self when i first attempted it

argh i dun get the other ones...

cm plz give us a hint for the 2nd one

 

[CM_Tutor]

Originally posted by :: ryan.cck ::
cisn@ - 1 = 2isin(n@/2)(cos(n@/2) - 2sin²(n@/2)
= isin(n@) + cosn@ - 1
= cisn@ -1

LHS = RHS

That is a really nice way of solving this - actually much better than the one I had in mind.

Originally posted by :: ryan.cck ::
cm plz give us a hint for the 2nd one

Hint: Describe the LHS of the result in (b) in two words.

 

[Faera]

for (b)

z + z^2 + z^3 + z^4 + .... + z^n = [z(z^n - 1)]/(z-1) ... sum of n terms in a geometric sequence.

[z(z^n - 1)]/(z-1) = (cis@{2i*sin(n@ / 2)*[cos(n@ /2) + isin(n@ / 2)]})/(cis@-1)

= [2i(sin(n@/2)(cis(2+n)@/2)]/{2i[sin(@/2)][cos(@/2) + isin(@/2)]

= [sin(n@/2)/sin(@/2)]{cis[(2@+n@-@)/2]}

= [sin(n@/2)/sin(@/2)][cis(n+1)@/2]

 

[CM_Tutor]

Faera, that's right. Everyone else, if you don't follow what Faera has done, try writing it out yourself, and add in some more steps.

(c) anyone?

 

[Faera]

(c)

cos@ + cos(2@) + cos(3@) + ... + cos(n@) = Re(z + z^2 + z^3 + ... + z^n)

= [sin(n@/2)]/[sin(@/2)] * [cos(n+1)@/2]

 

[CM_Tutor]

Originally posted by Faera
(c)

cos@ + cos(2@) + cos(3@) + ... + cos(n@) = Re(z + z^2 + z^3 + ... + z^n)

= [sin(n@/2)]/[sin(@/2)] * [cos(n+1)@/2]

You do mean [sin(n@ / 2) / sin(@ / 2)] * cos[(n + 1)@ / 2], don't you? If so, then yes, that's (c). Any progress on (d) and (e)?

 

[Faera]

.. 0.o ... oops. lol, yeah, i meant that.

i'm stuck on d... havent even loked at e yet...
geez... i should be memorizing my chem assessment... =P
this is too distracting -.-'