another complex number question (1 Viewer)

[insert-username]

Yeah, you're right. I've fixed that.


I_F

 

[Trev]

a) z=cosө+isinө
z⁶= (cosө+isinө)⁶ {let cosө=c and sinө=s)
=[6C0]c⁶i⁰s⁰+[6C1]c⁵i¹s¹+[6C2]c⁴i²s²+[6C3]c³i³s³+[6C4]c²i⁴s⁴+[6C5]c¹i⁵s⁵+[6C6]c⁰i⁶s⁶
Take out the real parts, which is equal to cos6ө and put it all in terms of cosө which will give you what you're supposed to show.

b) Similarly as sin6ө/cos6ө.

 

[pLuvia]

a) z=cosө+isinө
z⁶= (cosө+isinө)⁶ {let cosө=c and sinө=s)
=[6C0]c⁶i⁰s⁰+[6C1]c⁵i¹s¹+[6C2]c⁴i²s²+[6C3]c³i³s³+[6C4]c²i⁴s⁴+[6C5]c¹i⁵s⁵+[6C6]c⁰i⁶s⁶
Take out the real parts, which is equal to cos6ө and put it all in terms of cosө which will give you what you're supposed to show.

b) Similarly as sin6ө/cos6ө.

Whats [6C0] mean?

 

[pLuvia]

Take out the real parts, which is equal to cos6ө and put it all in terms of cosө which will give you what you're supposed to show.

what you mean?

 

[Trev]

tan3ө (i'll do this since it won't take as long)
z=cosө+isinө
z³=(cosө+isinө)³
= [3C0]cos³ө+[3C1]cos²өisinө+[3C2]cosөi²sin²ө+[3C3]i³sin³ө
= cos³ө-3cosө(1-cos²ө)+[3(1-sin²ө)sinө-sin³ө]i
= 4cos³ө-3cosө+(3sinө-4sin³ө)i
Equating real and unreal parts:
cos3ө = 4cos³ө-3cosө
sin3ө = 3sinө-4sin³ө
∴ tan3ө = (3sinө-4sin³ө)/(4cos³ө-3cosө)
= tanө[(3-4sin²ө)/(4cos²ө-3)]
Divide numerator and denominator by cos²ө {also; note that 1/cos²ө = 1+tan²ө}
= tanө[(3(1+tan²ө)-4tan²ө)/(4-3(1+tan²ө))]
= tanө[(3-tan²ө)/(1-3tan²ө)]
= (3tan²ө-tan³ө)/(1-3tan²ө) as required.

 

[pLuvia]

argh never mind hehe thanks trev, man trig identities you have to know them

 

[Trev]

Whats [6C0] mean?

Your teacher should have given you a quick run-down on binomial expansions.

Take out the real parts, which is equal to cos6ө and put it all in terms of cosө which will give you what you're supposed to show.

Since z=cosө+isinө the real part of z is cosө and the unreal part is sinө.
After z⁶ you will get real parts and unreal parts (unreal is the ones with 'i').
So taking out real parts of z⁶ is equal to cos6ө (by de moivres) which is equal to the real part of the expansion.

 

[pLuvia]

Your teacher should have given you a quick run-down on binomial expansions.


Since z=cosө+isinө the real part of z is cosө and the unreal part is sinө.
After z⁶ you will get real parts and unreal parts (unreal is the ones with 'i').
So taking out real parts of z⁶ is equal to cos6ө (by de moivres) which is equal to the real part of the expansion.

That looks like combinations? like in permutations and combinations?

 

[insert-username]

That looks like combinations? like in permutations and combinations?

The binomial expansion is based on combinations. It allows you to expand expressions in the form (x+y)ⁿ.


I_F

 

[sasquatch]

ive been doing ext 2 for a day now haha....today was my first day... and i did that first question and got

1- i +/- 2root(2i)

for my answer..

i noticed the first answerer did something with the -32i under the root and i guess thats something that i need to do too.. could someone please explain that??

edit: nevermind i just looked as something and yeah..

 

[Riviet]

........................... err, help please? I posted up a question the other day and i have not got a single response from anyone. I'm a little disappointed that you guys missed me and helped kadlil instead...

 

[Slidey]

I need some help with this question from the Advanced Mathematics textbook (Terry Lee):
If R is a real number, find the locus of z defined by
a) z=(1+iR)/(1-iR)
b) z=1/(1-iR)
c) z=iR/(1-iR)
I've only had a go at the first one, i substituted z=x+iy into a), then realised denominator, and figured out what x and y were in terms of R. This is where i didn't know what to do. The answers at the back of the book goes straight to x²+y²=[sub in x and y here]
which ends up as 1, so the locus turns out to be the unit circle. Please explain how? Thx. Don't worry about any working, the book has worked answers, i just don't get it.

c) (-R^2/[1+R^2],R/[1+R^2])
thus x=-y^2

Do the rest when sober.

 

[insert-username]

Riviet, yours wasn't something we'd just looked at in 2 unit Maths... mmm series


I_F

 

[Riviet]

Riviet, yours wasn't something we'd just looked at in 2 unit Maths... mmm series


I_F

What are you talking about?

 

[insert-username]

I was talking about kadil's complex number question that was in the form of a geometric series - I recognised that and so did Trev. I looked at yours earlier and thought "help".


I_F

 

[Riviet]

oh... i see.

 

[Trev]

Riviet:
It states that R is a real number, so after you have made R the subject, put z=x+iy and rationalised the denominator you take the real part of the answer (ignore the unreal part). This will be the required locus.

 

[Riviet]

in a), i get x=(1-R²)/(1+R²), y=2R/(1+R²)
Why and how do i know to go straight to x²+y²=1? I subed it in and it works right, but the reasoning?

 

[KFunk]

in a), i get x=(1-R²)/(1+R²), y=2R/(1+R²)
Why and how do i know to go straight to x²+y²=1? I subed it in and it works right, but the reasoning?

This isn't the most straight forward way to do it but what you've got is the same form as those t substitution things so if you let R = tan(@/2) then x = cos@ and y = sin@ ∴ x² + y² = 1. More a cool pattern than a method to use.

 

[Riviet]

Hmm... that soughta helps, thx KFunk