Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By johnsomeone

Math Help - Prove by induction

  1. #1
    Junior Member
    Joined
    Sep 2012
    From
    New York
    Posts
    55

    Prove by induction

    Use mathematical induction to prove the following theorem:
    (calculation of powers in C): Let z=r(cos θ + i sin θ) be any complex number and n be any positive integer. Then zn=rn(cos (nθ) + i sin (nθ)) or equivalently [r,θ]n=[rn, nθ].
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: Prove by induction

    Well, what's the base case?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2012
    From
    New York
    Posts
    55

    Re: Prove by induction

    what do you mean base case?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    146

    Re: Prove by induction

    Proof by induction goes:

    Write down the statement S you intend to prove, but think of it as a different statement for each different n. In that sense, think of your statement as a "function of n", S(n).

    Here, S(n) = "If z = r(\cos(\theta) + i \sin(\theta)), then z^n = r^n(\cos(n \theta) + i \sin(n \theta))."

    1st, prove S(0) is true.

    Here that means proving "If z = r(\cos(\theta) + i \sin(\theta)), then z^0 = r^0(\cos((0)(\theta)) + i \sin((0)(\theta)))".

    So do that.

    Next (and here's the sometimes confusing part) you need to prove a logical implication. You need to prove: "S(n) implies S(n+1) for integers n >= 0."

    You do that this way: ASSUME S(n) is true for some n>=0. Using that assumption, work some magic (i.e. do some algebra) to derive that S(n+1) must be true.

    (If you can do that, then you will have proven that S(n) is true for all non-negative integers n. You will have proven S by induction.)

    That looks like:
    -----------------------
    If z = r(\cos(\theta) + i \sin(\theta)), then z^n = r^n(\cos(n \theta) + i \sin(n \theta)) (this is our assumption S(n)).

    ... (work your algebraic magic here)...

    Therefore, z^{n+1} = r^{n+1}(\cos((n+1) \theta) + i \sin((n+1) \theta)) (this is our statement S(n+1)).
    -----------------------

    That's how you do it.

    You choose your "algebraic magic" by looking at where you are and where you want to go. And then you try to algebraically get there on the "easy side", and then typically do some hideous calculations on the "complicated side" which, if it's true and you don't make a mistake, will ultimately simplfy into the desired outcome.

    Here, that means since you have z^{n}, and you want to get to z^{n+1}, mutliply both sides of what you have by z:

    We have z^n = r^n(\cos(n \theta) + i \sin(n \theta)),

    so (z)(z^n) = (z)(r^n(\cos(n \theta) + i \sin(n \theta))),

    so z^{n+1} = (r(\cos(\theta) + i \sin(\theta))(r^n(\cos(n \theta) + i \sin(n \theta))).

    On the LHS, you now have your desired z^{n+1}, and on the RHS, you have a lot of algebra to do. But it does eventually work out to look like the desired final statement =S(n+1).
    Last edited by johnsomeone; September 27th 2012 at 12:29 AM.
    Thanks from MarkFL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591

    Re: Prove by induction

    to give you an idea of "how you might proceed", let's look at the case n = 2 (from this, we might get an idea of how to go from S(n) to S(n+1)):

    suppose z = r(\cos(\theta) + i\sin(\theta)).

    then z^2 = [r(\cos(\theta) + i\sin(\theta))]^2 = r^2(\cos(\theta) + i\sin(\theta))^2

    = r^2[\cos^2(\theta) - \sin^2(\theta) + i(2\sin(\theta)\cos(\theta))] (using the rules of complex multiplication)

    = r^2(\cos(2\theta) + i\sin(2\theta))

    it appears the angle addition formulae for sine and cosine may be of use.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  2. Prove each of the following by induction
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: February 25th 2010, 10:45 AM
  3. Prove by induction
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: February 18th 2010, 07:18 AM
  4. Prove by induction
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 10th 2009, 08:11 AM
  5. need help with induction prove..
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: February 16th 2008, 09:13 AM

Search Tags


/mathhelpforum @mathhelpforum