complex no (1 Viewer)

[bobo123]

a friend sent me this:

part (i)
show that:
sum of series

n
E (z+z^2+......z^k) = nz/(1-z) - z^2(1-z^n)/(1-z)^2
k=1

part (ii)
let z= cos@+isin@, where 0<@<2pi

deduce that

n
E (sin@+sin2@...........+sink@)
k=1

= ((n+1)sin@ - sin(n+1)@)/4sin^2 (@/2)


you may assume that

z/(1-z) = i(cis(@/2))/2sin(@/2)



i got very close but.............yeah

 

[spice girl]

where did you get upto?

 

[KeypadSDM]

I like alot, very nice question
it's a change from the trials i'm doing now...

 

[bobo123]

i tried muscling it out but cant seem to get the exact answer
i was just hoping for someone here to find a shortcut

 

[McLake]

Well part i) is a geometric progression.

Working on part ii) ...

 

[bobo123]

im thinking its just my carelessness screwing me over

which question do you reckon this one would be in the paper? difficulty wise (1-8)

 

[Mathematician]

haha

haha its q8 from 2000 school certificate.

 

[bobo123]

school certificate?
are you really that powerful in maths?

 

[spice girl]

You're supposed to use


Im{nz/(1-z) - z^2(1-z^n)/(1-z)^2} =
n
E (sin@+sin2@...........+sink@)
k=1

Plug z/(1-z) = i(cis(@/2))/2sin(@/2) in, and expand and simplify.

 

[Mathematician]

...

haha sorry, u know i meant hsc

 

[bobo123]

yes i got it now
stupid stupid careless mistakes