Let z = 2 + i7 and let w = 3 − i8. Find zw,z + w,z2, and w∕z.
Give the complete solution to x4 + 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
De Moivre’s theorem says
[r (cost+ isin t)]
n = rn
(cosnt+ isin nt)
for n a positive
integer. Does this formula continue to hold for all integers, n, even negative integers?
You already know formulas for cos
(x + y)
and these were used to
prove De Moivre’s theorem. Now using De Moivre’s theorem, derive a formula for
and one for cos
. Hint:Use the binomial theorem.
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 x3 + 8 as a product of linear factors.
Write x3 + 27 in the form
(x2 + ax+ b)
where x2 + ax + b cannot be factored
any more using only real numbers.
Completely factor x4 + 16 as a product of linear factors.
Factor x4 + 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
zm. Also verify that ∑k=1mzk = ∑k=1mzk. 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.
= anxn + an−1xn−1 +
+ a1x + a0 where all the ak are real numbers.
Suppose also that p
= 0 for some z ∈ ℂ. Show it follows that p
= 0 also.
I claim that 1 = −1. Here is why: −1 = i2 =
= 1. 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
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
Show that ℂ cannot be considered an ordered field. Hint:Consider i2 = −1. Recall that
1 > 0 by Proposition 1.4.2.
Say a + ib < x + iy if a < x or if a = x, then b < y. This is called the lexicographic order.
Show that any two different complex numbers can be compared with this order. What goes
wrong in terms of the other requirements for an ordered field.
With the order of Problem 18, consider for n ∈ ℕ the complex number 1 −
. Show that
with the lexicographic order just described, each of 1 − in is an upper bound to all these
numbers. Therefore, this is a set which is “bounded above” but has no least upper bound
with respect to the lexicographic order on ℂ.