- Prove by induction that ∑
_{k=1}^{n}k^{3}=n^{4}+n^{3}+n^{2}. - Prove by induction that whenever n ≥ 2,∑
_{k=1}^{n}>. - Prove by induction that 1 + ∑
_{i=1}^{n}i=! . - The binomial theorem states
^{n}= ∑_{k=0}^{n}x^{n−k}y^{k}whereProve the binomial theorem by induction. Next show that

- Let z = 5 + i9. Find z
^{−1}. - Let z = 2 + i7 and let w = 3 − i8. Find zw,z + w,z
^{2}, and w∕z. - Give the complete solution to x
^{4}+ 16 = 0. - Graph the complex cube roots of 8 in the complex plane. Do the same for the four fourth roots of 16. ▸
- If z is a complex number, show there exists ω a complex number with = 1 and ωz =.
- De Moivre’s theorem says
^{n}= r^{n}for n a positive integer. Does this formula continue to hold for all integers n, even negative integers? Explain. ▸ - You already know formulas for cosand sinand these were used to prove De Moivre’s theorem. Now using De Moivre’s theorem, derive a formula for sinand one for cos. ▸
- If z and w are two complex numbers and the polar form of z involves the angle θ while the polar form
of w involves the angle ϕ, show that in the polar form for zw the angle involved is θ + ϕ. Also, show
that in the polar form of a complex number z, r = .
- Factor x
^{3}+ 8 as a product of linear factors. - Write x
^{3}+ 27 in the formwhere x^{2}+ ax + b cannot be factored any more using only real numbers. - Completely factor x
^{4}+ 16 as a product of linear factors. - Factor x
^{4}+ 16 as the product of two quadratic polynomials each of which cannot be factored further without using complex numbers. - If z,w are complex numbers prove zw = zw and then show by induction that ∏
_{j=1}^{n}z_{j}= ∏_{j=1}^{n}z_{j}. Also verify that ∑_{k=1}^{m}z_{k}= ∑_{k=1}^{m}z_{k}. In words this says the conjugate of a product equals the product of the conjugates and the conjugate of a sum equals the sum of the conjugates. - Suppose p= a
_{n}x^{n}+ a_{n−1}x^{n−1}++ a_{1}x + a_{0}where all the a_{k}are real numbers. Suppose also that p= 0 for some z ∈ ℂ. Show it follows that p= 0 also. - Show that 1 + i,2 + i are the only two zeros to
so the zeros do not necessarily come in conjugate pairs if the coefficients are not real.

- I claim that 1 = −1. Here is why.
This is clearly a remarkable result but is there something wrong with it? If so, what is wrong?

- De Moivre’s theorem is really a grand thing. I plan to use it now for rational exponents, not just
integers.
Therefore, squaring both sides it follows 1 = −1 as in the previous problem. What does this tell you about De Moivre’s theorem? Is there a profound difference between raising numbers to integer powers and raising numbers to non integer powers?

- Review Problem 10 at this point. Now here is another question: If n is an integer, is it always true
that
^{n}= cos− isin? Explain. - Suppose you have any polynomial in cosθ and sinθ. By this I mean an expression of the form
∑
_{α=0}^{m}∑_{β=0}^{n}a_{αβ}cos^{α}θ sin^{β}θ where a_{αβ}∈ ℂ. Can this always be written in the form ∑_{γ=−(n+m ) }^{m+n}b_{γ}cosγθ + ∑_{τ=−(n+m ) }^{n+m}c_{τ}sinτθ? Explain. - Show that ℂ cannot be considered an ordered field. Hint: Consider i
^{2}= −1. - Suppose p= a
_{n}x^{n}+ a_{n−1}x^{n−1}++ a_{1}x + a_{0}is a polynomial and it has n zeros,listed according to multiplicity. (z is a root of multiplicity m if the polynomial f

=^{m}divides pbutfdoes not.) Show that - Give the solutions to the following quadratic equations having real coefficients.
- x
^{2}− 2x + 2 = 0 - 3x
^{2}+ x + 3 = 0 - x
^{2}− 6x + 13 = 0 - x
^{2}+ 4x + 9 = 0 - 4x
^{2}+ 4x + 5 = 0

- x
- Give the solutions to the following quadratic equations having complex coefficients. Note
how the solutions do not come in conjugate pairs as they do when the equation has real
coefficients.
- x
^{2}+ 2x + 1 + i = 0 - 4x
^{2}+ 4ix − 5 = 0 - 4x
^{2}+x + 1 + 2i = 0 - x
^{2}− 4ix − 5 = 0 - 3x
^{2}+x + 3i = 0

- x
- Prove the fundamental theorem of algebra for quadratic polynomials having coefficients in ℂ. That is,
show that an equation of the form ax
^{2}+ bx + c = 0 where a,b,c are complex numbers, a≠0 has a complex solution. Hint: Consider the fact, noted earlier that the expressions given from the quadratic formula do in fact serve as solutions. - Prove the Euclidean algorithm: If m,n are positive integers, then there exist integers q,r ≥ 0 such
that r < m and
Hint: You might try considering

and picking the smallest integer in S or something like this. It was done in the chapter, but go through it yourself.

- Recall that two polynomials are equal means that the coefficients of corresponding powers of λ are
equal. Thus a polynomial equals 0 if and only if all coefficients equal 0. In calculus we usually think
of a polynomial as 0 if it sends every value of x to 0. Suppose you have the following
polynomial
where it is understood to be a polynomial in ℤ

_{2}. Thus it is not the zero polynomial. Show, however, that this equals zero for all x ∈ ℤ_{2}so we would be tempted to say it is zero if we use the conventions of calculus. - Prove Wilson’s theorem. This theorem states that if p is a prime, then ! + 1 is divisible by p. Wilson’s theorem was first proved by Lagrange in the 1770’s. Hint: Check directly for p = 2,3. Show that p − 1 = −1 and that if a ∈, then
^{−1}≠a. Thus a residue class a and its multiplicative inverse for a ∈occur in pairs. Show that this implies that the residue class of! must be −1. From this, draw the conclusion. - Show that in the arithmetic of ℤ
_{p},^{p}=^{p}+^{p}, a well known formula among students. - Consider ∈ ℤ
_{p}for p a prime, and suppose≠1,0. Fermat’s little theorem says that^{p−1}= 1. In other words^{p−1}− 1 is divisible by p. Prove this. Hint: Show that there must exist r ≥ 1,r ≤ p− 1 such that^{r}= 1. To do so, consider 1,,^{2},. Then these all have values in, and so there must be a repeat in, say p − 1 ≥ l > k and^{l}=^{k}. Then tell why^{l−k}−1 = 0. Let r be the first positive integer such that^{r}= 1. Let G =. Show that every residue class in G has its multiplicative inverse in G. In fact,^{k}^{r−k}= 1. Also verify that the entries in G must be distinct. Now consider the sets bG ≡where b ∈. Show that two of these sets are either the same or disjoint and that they all consist of r elements. Explain why it follows that p− 1 = lr for some positive integer l equal to the number of these distinct sets. Then explain why^{p−1}=^{lr}= 1. - Let pand qbe polynomials. Then by the division algorithm, there exist polynomials l, requal to 0 or having degree smaller than psuch that
If k

is the greatest common divisor of pand q, explain why kmust divide r. Then argue that kis also the greatest common divisor of pand r. Now repeat the process for the polynomials pand r. This time, the remainder term will have degree smaller than r. Keep doing this and eventually the remainder must be 0. Describe an algorithm based on this which will determine the greatest common divisor of two polynomials. - Consider ℤ
_{m}where m is not a prime. Show that although this will not be a field, it is a commutative ring with unity. - This and the next few problems are to illustrate the utility of the limsup. A sequence of numbers
in ℂ is called a Cauchy sequence if for every ε > 0 there exists m such that if k,l ≥ m, then< ε. The complex numbers are said to be complete because any Cauchy sequence converges. This is one form of the completeness axiom. Using this axiom, show that ∑
_{k=0}^{∞}r^{k}≡ lim_{n→∞}∑_{k=0}^{n}r^{k}=whenever r ∈ ℂ and< 1. Hint: You need to do a computation with the sum and show that the partial sums form a Cauchy sequence. - Show that if ∑
_{j=1}^{∞}converges, meaning that lim_{n→∞}∑_{j=1}^{n}exists, then ∑_{j=1}^{∞}c_{j}also converges, meaning lim_{n→∞}∑_{j=1}^{n}c_{j}exists, this for c_{j}∈ ℂ. Recall from calculus, this says that absolute convergence implies convergence. - Show that if ∑
_{j=1}^{∞}c_{j}converges, meaning lim_{n→∞}∑_{j=1}^{n}c_{j}exists, then it must be the case that lim_{n→∞}c_{n}= 0. - Show that if limsup
_{k→∞}^{1∕k}< 1, then ∑_{k=1}^{∞}converges, while if limsup_{n→∞}^{1∕n}> 1, then the series diverges spectacularly because lim_{n→∞}fails to equal 0 and in fact has a subsequence which converges to ∞. Also show that if limsup_{n→∞}^{1∕n}= 1, the test fails because there are examples where the series can converge and examples where the series diverges. This is an improved version of the root test from calculus. It is improved because limsup always exists. Hint: For the last part, consider ∑_{n}and ∑_{n}. Review calculus to see why the first diverges and the second converges. - Consider a power series ∑
_{n=0}^{∞}a_{n}x^{n}. Derive a condition for the radius of convergence using limsup_{n→∞}^{1∕n}. Recall that the radius of convergence R is such that if< R, then the series converges and if> R, the series diverges and if= R is it not known whether the series converges. In this problem, assume only that x ∈ ℂ. - Show that if a
_{n}is a sequence of real numbers, then liminf_{n→∞}= −limsup_{n→∞}a_{n}.

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