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)]

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Best answer

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.

by Top Rated User (710k points)
selected by

Related questions

1 answer
1 answer
asked Feb 24, 2016 in Algebra 2 Answers by karan91 Level 2 User (1.3k points) | 134 views
1 answer
1 answer
1 answer
1 answer
1 answer
asked Feb 16, 2016 in Algebra 2 Answers by karan91 Level 2 User (1.3k points) | 111 views
1 answer
1 answer
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
84,401 questions
89,210 answers
1,988 comments
7,485 users