Complex numbers question (1 Viewer)

[thoth1]

It is given that for 0 < arg z< pi/2

Show that and express z-1 in mod arg form

Hence show that

i did it but my way seem over the top.

 

[kooliskool]

Hi,

I did your z+1 before in this thread: http://community.boredofstudies.org/showthread.php?t=261664, then I am guessing the z-1 will be similar.

Good luck with the rest.

 

[thoth1]

thanks kooliskool. thats the method i used as well. the algebra just seemed a little heavy for a question 2 so thats why i posted.

 

[kooliskool]

Oh, ic, well, that's the only way I can think of, I don't think there's any faster way already.

 

[clementc]

Hey thoth,

There's another way you can do it, but it only works out really nicely for this question. I think if it was something else it wouldn't work. But here goes!

So we know...










And that's it! (for the first part anyway - the z-1 is similar probably with a chunk of imaginary stuff or two =D )
God that took so much incredibly longer to type than it took to actually work it out LOL
Anyway good luck studying! Packing my bag...PARTY/BLUDGE TIME TOMORROW YAY

 

[largarithmic]

You can look at this geometrically as well. If O is the origin, A represents z, and B represents z+1, and C represents 1, then because z has unit modulus (implied by z = cistheta), and because then AB has length 1 (try tail-to-tip addition or whatever) then OAB is isosceles. Another way of seeing this would be, by the rules of vector addition, OABC is a parallelogram, and because OA = 1 = OC then OABC is a rhombus.

It then follows that angle COB = half angle COA = theta/2, and also you can then use the cosine or sine rule (whatever suits you best) to get that OB = 2cos theta/2. Combining those together you know that the argument of z+1 is theta/2, and the modulus has to be 2cos theta/2, so you get z+1 = 2costheta/2 cistheta/2.

A lot of random complex numbers questions like this can be done geometrically, or at least motivated geometrically. A golden rule I reckon of compex numbers is just DRAW A DIAGRAM; even if youre not gonna use geometry, having a picture in front of you will almost always help.

 

[math man]

i will let theta equal x to make it simply for me

z = cosx + isinx,

z + 1 = cosx + 1 + isinx, to eliminate the 1 use cosx = 2cos^2(x/2) - 1 and also sub sinx = 2sin(x/2)cos(x/2)

z + 1 = 2cos^2(x/2) - 1 + 1 + 2isin(x/2)cos(x/2) ...now we can factor cos(x/2) out and the first part is done

z + 1 = 2cos(x/2)( cos(x/2) + isin(x/2) ) ...for z - 1 we will sub cosx = 1 - 2sin^2(x/2) to eliminate the one

z - 1 = 1 - 2sin^2(x/2) - 1 + 2isin(x/2)cos(x/2) ....we factor -2sin(x/2) out

z - 1 = -2sin(x/2)( sin(x/2) - icos(x/2) ) ...now we divide (z-1) by (z + 1)
(z-1)/(z+1) = [-2sin(x/2){sin(x/2) - icos(x/2)}]/[2cos(x/2){cos(x/2) + sin(x/2)}] ...we now times by conjugate of denom and noticing the denom goes to one...

= (-tan(x/2))[sin(x/2) - icos(x/2)][(cos(x/2) - isin(x/2)] ....now we expand this and something great happens

= (-tan(x/2))[sin(x/2)cos(x/2) -isin^2(x/2) - icos^2(x/2) - cos(x/2)sin(x/2)] ...and we notice the real parts cancel, hence the real part of (z-1)(z+1) is 0
hope this helps

 

[AAEldar]

hope this helps

I had to clean that up >.<

Must have been hard for you to follow what you were writing though, I'll give you credit for that!

 

[math man]

I had to clean that up >.<

Must have been hard for you to follow what you were writing though, I'll give you credit for that!

lol thankyou...i didnt realise the latex thing till you said it on the other post