Prove by induction

**franios** · September 26th 2012, 06:18 PM

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].

**Prove It** · September 26th 2012, 06:28 PM

Well, what's the base case?

**franios** · September 26th 2012, 06:31 PM

what do you mean base case?

**johnsomeone** · September 27th 2012, 12:21 AM

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).

**Deveno** · September 27th 2012, 03:23 PM

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.

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