Prove, by contradiction, that if w,z∈C such that |w|≤1, and (w^n)z+(w^n+1)z^2+.....+(wz)^n =1, then |z|>(1/2).
in Algebra 2 Answers by Level 2 User (1.3k points)

Do you mean: (w^n)z+(w^(n-1))z^2+...+wz^n=1?

Or, effectively, w^nz[1+wz+(wz)^2+...(wz)^(n-1)]=1?

This would make the general term: (w^(n-r))z^(r+1) or (w^(n+r))(z^(r+1)), where r is between 0 and n-1 (0<r<n), where n is assumed to be an integer.

Yes.... I'm sorry... That was a mistake Can you give a final answer?? [prove |z|>(1/2)]

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Given the validity of the comment attached to the question, and it's resolution, we can take a look at the implication of the question. To be true for w and z generally, we can take a specific case to demonstrate the argument.

Let w=0.8+0.6i, which satisfies the requirement |w|=1, a special case of |w|<1.

For any value of n, the binomial expansion of (a+b)^n=1 only when a+b=1, i.e., b=1-a.

The binomial expansion is (a+b)^n=a^n+nba^(n-1)+(n(n-1)/2)b^2a^(n-2)+...+b^n=1.

The general term is (n!/(r!(n-r)!))a^(n-r)b^r, where 0<r<n. This can also be written (nCr)a^(n-r)b^r, where nCr is the combination function (the number of ways of combining r different objects from a set of n such objects).

The resembles the given expansion of w and z, but it has additional coefficients. Let z=1-w. Does the expansion hold when w and z are complex?

With the chosen value of w, z=1-(0.8+0.6i)=0.2-0.6i. |z|=sqrt(0.04+0.36)=sqrt(0.40)=0.6325 approx. This is greater than 1/2.

Now, let w=0.3+0.4i, |w|=0.5, z=1-(0.3+0.4i)=0.7-0.4i, |z|=0.8062 approx>1/2.

Let w=0.03+0.04i, |w|=0.05, z=1-(0.03+0.04i)=0.97-0.04i, |z|=0.9708 approx>1/2.

As |w| gets smaller, |z| gets larger, so moving further away from 1/2 and towards 1.

What is w+z? In all of the above cases w+z=1, by definition, and so the binomial expansion also =1.

Just seen your comment.

Let w=0.96+0.28i, |w|=1, z=0.04-0.28i, |z|=0.2828 approx, which is <1/2. So the proposition would not be true for the binomial expansion of (w+z)^n=1 for all n. That means we require a different expression than w+z. However, we do know that (...)^n must be 1 for all n and we need to find an alternative expression similar to the one given.

Let w=a+isqrt(p^2-a^2), |w|=sqrt(a^2+p^2-a^2)=p<1. This is the general form for w. In the above, z=1-a-isqrt(p^2-a^2) and |z|=sqrt((1-a)^2+p^2-a^2)=sqrt(1-2a+a^2+p^2-a^2)=sqrt(1-2a+p^2). When p=1, this is sqrt(2(1-a)) which can be less than 1/2 (7/8<a<1). This explains why |z| can be less than 1/2.

So, how can we find the right expression? Let n=1. We know the expansion has to be true for all n, so picking one specific value should still work. If we go back to the original question, we would have wz=1. For n=2, w^2z+wz^2=1; wz^2+w^2z-1=0 is a quadratic with solutions: z=(-w^2+sqrt(w^4+4w))/2w. That doesn't look promising!

I have to say that we are now stuck until we have the right polynomial expansion to work from. However, there is a solution! If wz=1 and w+z=1, then w^2-w+1=0 and w=1/2+isqrt(3)/2, z=1/2-isqrt(3)/2. |w|=|z|=1 (<1 and >1/2) and wz/(w+z)=1 so ((w+z)/wz)^n=((1/z)+(1/w))^n=(wz/(w+z))^n=1. ((1/z)+(1/w))^n=(w+z)^n=1. Therefore, w and z are particular complex numbers for which the proposition is true, and the binomial expansion applies.

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