n00bie complex number q's (1 Viewer)

[CM_Tutor]

Originally posted by ...
ahhh..so u don't just sub K as 0,1..but rather subing the mod in there as well?

Sorry, I'm not following your question. Could you rephrase / restate / clarify? Thanks.

 

[...]

hahaha..err..ok, but be careful, u'll probably die laughting at my stupidity..

By De Moivre's Theorem, the n distinct values of z are:

cos(2k.pi/n) + isin(2k.pi/n)
for k = 0,1,2,...,n-1


i just dun really understand the steps:

Express -i in mod-arg form. You should get cos(-pi/2) + isin(-pi/2)

to..
z = cos(k * pi - pi/4) + isin(k * pi - pi/4)

 

[Ragerunner]

You don't just pick k = 0,1,2,3,4,5...n

You only use the values of k that lie between (-pi, pi]

I think. Could be wrong. *wait's to be proved wrong*

 

[:: ck ::]

You only use the values of k that lie between (-pi, pi]

*lost*
then agn i didn read the questoin properly

u can use anything for k cant u? easiest to use ones which are negative of eachtoher eg : 0, -1 , 1, -2, 2 etc...

editz : i think ur talking abt principle argument? .. anyways.. if its not between -pi and pi u can always just -2pi until it lies in principle argument

 

[CM_Tutor]

Originally posted by ...
hahaha..err..ok, but be careful, u'll probably die laughting at my stupidity..

Don't be afraid to ask questions - asking a question to help improve your understanding is never a stupid thing to do. In any case, I understand the problem now, so let me explain further.

I started out by saying to express -i in mod-arg form. This you do by the method describe by martin310015 above, ie:

|-i| = 1

arg(-i) = - arg(i) = -pi / 2

So, -i = cos(-pi/2) + isin(-pi/2)

In the statement that you quoted, I went on to say:
So, z² = cos(-pi/2) + isin(-pi/2) = cos(2k * pi - pi/2) + isin(2k * pi - pi/2), where k is any integer.

This is a statement about the values of z², not the values of z, and I have not (at that point) applied De Moivre's Theorem. This I do in going to the next line, which I'll now do showing all the working:

z² = cos(2k * pi - pi/2) + isin(2k * pi - pi/2), where k is any integer
So, by square rooting, z = [cos(2k * pi - pi/2) + isin(2k * pi - pi/2)]^1/2
= cos[(2k * pi - pi/2) * (1 / 2)] + isin[(2k * pi - pi/2) * (1 / 2)], applying De Moivre's theorem
= cos[(2k * pi) * (1 / 2) - (pi / 2) * (1 / 2)] + isin[(2k * pi) * (1 / 2) - (pi / 2) * (1 / 2)]
= cos(k * pi - pi/4) + isin(k * pi - pi/4)

We can now take any integer values of k we like, but we only need to take two consecutive values, as all others will just be repeats of previous answers. I chose k = 0 and k = 1 for convenience, but you could have used any to consecutive integers.

This then produces the two answers z = cos(-pi/4) + isin(-pi/4) = (1 / sqrt(2)) + i * (-1 / sqrt(2)) = (1 - i) / sqrt(2)
and z = cos(3pi/4) + isin(3pi/4) = (-1 / sqrt(2)) + i * (1 / sqrt(2)) = (-1 + i) / sqrt(2).

Ask again if this still isn't clear

Ragerunner, k can be any integer, and you are getting confused about principal arguments. We usually try to choose k values such that the argument satisifies (-pi, pi] - indeed, this is why I took k = 0 and k = 1 - but it is the argument - in this case (k * pi - pi/4) - that we restrict, not k. As ryan.cck has noted, you can always adjust to be a principal argument later, if you choose values of k poorly.

 

[...]

Originally posted by CM_Tutor
z² = cos(2k * pi - pi/2) + isin(2k * pi - pi/2), where k is any integer

hmm..
i understood everything..cept that line

z² = cos(-pi/2) + isin(-pi/2) = cos(2k.pi - pi/2) + isin(2k.pi - pi/2)



like where did 2k.pi come from???
lol...sorry, i'm learning complex numbers all myself atm

 

[nike33]

the 2k.pi is just the general solution

 

[:: ck ::]

yup rotate a pt by 2kpi radians u get the same point

 

[CM_Tutor]

Originally posted by ...
hmm..
i understood everything..cept that line

z² = cos(-pi/2) + isin(-pi/2) = cos(2k.pi - pi/2) + isin(2k.pi - pi/2)



like where did 2k.pi come from???
lol...sorry, i'm learning complex numbers all myself atm

Nike33 and ryan.cck are correct, all I have done is generalise from the principal argument to all possible arguments. This is something that you will do quite often. If you aren't convinced that it is true, you can expand to confirm. ie:

cos(2k * pi - pi/2) + isin(2k * pi - pi/2)
= [cos(2k * pi) * cos(pi / 2) + sin(2k * pi) * sin (pi / 2)] + i[sin(2k * pi) * cos(pi / 2) - cos(2k * pi) * sin(pi / 2)]
= [1 * cos(pi / 2) + 0 * sin(pi / 2)] + i[0 * cos(pi / 2) - 1 * sin(pi / 2)]
= cos(pi / 2) - isin(pi / 2)
= cos(-pi / 2) + isin(-pi / 2), as required, noting that cos@ is even and sin@ is odd

 

[...]

ah..

yeap yeap, i understand it now ^^

thanks everyone

now heres another noobie one

say i were to evaulate this
(3 + 4i)⁵
is there a shorter method or do i have to take it all out :\

 

[ND]

I'm not sure what you mean by "take it all out" but i'd do it by converting to mod arg and using de moive's. Should come out to be 5^5*cis(5*arctan(4/3)).

Note: cis@ means cos@ + isin@ (us 4u people are lazy)

 

[:: ck ::]

i think he meant by binomial expansion =\

demoivre's is the way to go..

 

[...]

but what happens if i was told to put it in a + bi form?

and i thought u can use de movires theorm when u have sin@ and cos@
(which in this case it doesn't)

 

[ND]

If it's to be put in a+ib form then just expand out. Haven't they taught you how to convert betweem rectangular and polar forms? a+ib=sqrt(a^2+b^2)(cos(arctan(b/a)+isin(arctan(b/a))).

 

[:: ck ::]

mmm when u have something in form rcis@ u want to put it into rectangular form

expand out to rcos@ + risin@

use values for wot cos@ and sin@ and r is equal to... expand out... and it will b in rectangular form...

 

[nike33]

... get fitz 4u its very simple / concise and would give you good foundations

 

[...]

Originally posted by ND
If it's to be put in a+ib form then just expand out. Haven't they taught you how to convert betweem rectangular and polar forms? a+ib=sqrt(a^2+b^2)(cos(arctan(b/a)+isin(arctan(b/a))).

yeap they have..

Originally posted by nike33
... get fitz 4u its very simple / concise and would give you good foundations

i'll ask my friend, he still has his...

 

[...]

hmm..this is a big wtf

Use De Moivre's theorem to show:
zⁿ + z^-n = 2.cos(n@) Key: @ = theta)

 

[ND]

That's when |z|=1.

Ok let z=cos@+isin@.
z^n+z^-n=(cos@+isin@)^n+(cos@+isin@)^-n
=cos(n@)+isin(n@)+cos(-n@)+isin(-n@) (by de moivre's)
cos(n@)+isin(n@)+cos(n@)-isin(n@) (because cos(-@)=cos@ and sin(-@)=-sin@)
=2cos(n@)

 

[...]

oh, thats easy
should have informed me that z = cos@ + isin@

or is that taken for granted in complex number?