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Complex number theorem help

  1. Sep 26, 2011 #1
    Hi guys, just before i ask this question i would like to let you know that i am a year 11 student, who has decided to study next years Specialist math (highest level of maths) course early to get a head start as i am nervous for next year(year 12.) :)

    1. The problem statement, all variables and given/known data
    2zzkn51.jpg

    and

    bhypt1.jpg

    The first one i understand part A and B i have attempted C but am not sure if it the correct method and D i have attempted but have not found an answer, i believe they are related to the second picture. I do not understand how that formula works and need help understanding it, what is the purpose of it?


    2. Relevant equations

    bhypt1.jpg and cisӨ=CosӨ+isinӨ

    3. The attempt at a solution

    z =2cisӨ
    A) |z| = 2
    arg z = Ө

    B)
    _
    Z = 2(cosӨ-isinӨ)
    = 2cosӨ-isinӨ
    = 2cos(-Ө)+isin(-Ө)
    = 2cis(-Ө)

    C) -z = -2cisӨ
    = -2(cosӨ+isinӨ)
    = -2cosӨ-isinӨ
    = -2cosӨ+isin(-Ө)
    From here i am stuck, the textbook shows the answer as 2cis(Ө+∏)

    D)
    _
    -Z = -2(cosӨ-isinӨ)
    = -2cosӨ+2isin(-Ө)
    From here i am stuck again, the textbook shows the answer as 2cis( ∏-Ө)

    I believe the answers are used using the formula stated above cis(Ө+k2∏)=cisӨ but i do not understand what it means or why it is used

    Please remember that i am self-teaching myself this subject at the moment and will be undertaking it next year for year 12.

    Thanks!! :)
     
  2. jcsd
  3. Sep 26, 2011 #2

    Mentallic

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    I took the same class as you, and it wasn't so bad. If you work at it (a couple of hours of study a night, each night) you'll probably be one of the best in the class in no time! Complex numbers was first though and it did scare quite a few students off, but trust me, it's nothing to get scared of.

    Have you heard of the argand diagram? The formula you're given would make perfect sense on it. Basically on its own it's not very helpful, but where it becomes useful is in finding roots of polynomials later on. For example, x3=1, the solution is 1, right? Well, there is a theorem that states that for an nth degree polynomial, there are n roots (but not all are necessarily distinct, which means they're not all different). There are in fact 3 roots in that equation, and you could solve it by factoring, such as
    x3-1=0
    (x-1)(x2+x+1)=0

    Then solve each factor, but there is also another way using the polar form of complex numbers that can quickly and easily find the roots of x3=-1, and when you learn about that you'll need to use the [itex]+2k\pi[/itex] at the end.


    Ok enough about that, for

    c)
    Again the argand diagram would be helpful in knowing what the answer should be, but I'll assume you don't. In that case you need to use the properties of the trig functions, such as how you earlier used -sin(x)=sin(-x)
    Expand [itex]\cos(\pi+x)[/itex] and [itex]\sin(\pi+x)[/itex]. See what you get.

    d)
    Again it's similar to c. There are some relations between trigs such as [itex]\sin(\pi/2-x)=cos(x)[/itex] which are quite useful in certain situations such as these.
     
  4. Sep 26, 2011 #3
    EDIT: sorry for edit, i did the question wrong again, but i found that cos(pi+x)=cos(pi-x)=-cosx and that sin(pi+x)=-sinx=sin(-x) my working out for C is...

    -z=-2cisx
    =-2cos(x)-isin(x)
    = 2cos(x)+isin(-x)
    =2cos(pi+x)+2isin(pi+x)
    = 2cis(pi+x)
    Ill attempt D and edit when im done thanks! :))

    D)

    _
    -z= -2cos(pi+x)-2isin(pi+x)
    = 2cos(pi-x)+2isin(pi+x) (as cos(pi+x)=cos(pi-x) ) and (sin(pi+x)=-sinx=sin(-x)
    = 2cis(pi-x)
    Hows that, Mentallic? :)
     
    Last edited: Sep 26, 2011
  5. Sep 26, 2011 #4

    Mentallic

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    Fixed and done! :smile:
     
  6. Sep 26, 2011 #5
    yeyy!!! :)) does that come up a lot the use of trig in complex numbers? do you mind if i ask you another quick question?

    I know how to solve this following question but i dont know what radians to use for 1582nt0.jpg

    Why do they use 3pi/12 and 4pi/12, why not 11pi/12 and pi/12 ? they all add up to the same value, do they all work the same or is their a rule to which raidans to use?

    For example what radians would we use with this and why use a specific amount of radians rather then another group of radians or does it not matter? cos(11pi/12), sin(11pi/12)
    to find in simplest form.

    Thanks, sorry to bother! :)
     
  7. Sep 26, 2011 #6

    Mentallic

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    Because there are some angles that give "nice" results.

    [tex]\cos\frac{3\pi}{12}=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}[/tex]

    and

    [tex]\cos\frac{4\pi}{12}=\cos\frac{\pi}{3}=\frac{1}{2}[/tex]

    Similarly for sin(x), we know how to calculate it quickly and easily if the angle of x is any of the following:

    [tex]0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{2}[/tex]

    and any integer multiples of those.

    The question is interesting because it gives us an exact answer to [itex]\cos\frac{7\pi}{12}[/itex] and [itex]\sin\frac{7\pi}{12}[/itex] which we wouldn't have known otherwise. If you used other angles such as [itex]\pi/12[/itex] and [itex]\pi/2[/itex] (which adds to [itex]7\pi/12[/itex]) then we don't know how to evaluate the cosine and sine of [itex]\pi/12[/itex] so it's pretty useless.

    By the way, you couldn't use 11pi/12 and pi/12 anyway because they don't add to [itex]7\pi/12[/itex]
     
  8. Sep 27, 2011 #7
    Okay i understand now so prety much just the simplest ones which i can evaluate using the unit circle or general knowledge, i was using my calculator to evaluate. I don't know why i said 11pi/12 and pi/12 must have read the question wrong.

    Thank you very much for your assistance it has helped me a lot!!

    Thanks! :)
     
  9. Sep 27, 2011 #8

    Mentallic

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    You're welcome :smile:
     
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