- 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. - 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.

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