2.10. EXERCISES 29

17. If z,w are complex numbers prove zw= zw and then show by induction that ∏nj=1 z j =

∏nj=1 z j. Also verify that ∑

mk=1 zk = ∑

mk=1 zk. In words this says the conjugate of a

product equals the product of the conjugates and the conjugate of a sum equals thesum of the conjugates.

18. Suppose p(x) = anxn +an−1xn−1 + · · ·+a1x+a0 where all the ak are real numbers.Suppose also that p(z) = 0 for some z ∈ C. Show it follows that p(z) = 0 also.

19. Show that 1+ i,2+ i are the only two zeros to

p(x) = x2− (3+2i)x+(1+3i)

so the zeros do not necessarily come in conjugate pairs if the coefficients are not real.

20. I claim that 1 =−1. Here is why.

−1 = i2 =√−1√−1 =

√(−1)2 =

√1 = 1.

This is clearly a remarkable result but is there something wrong with it? If so, whatis wrong?

21. De Moivre’s theorem is really a grand thing. I plan to use it now for rational expo-nents, not just integers.

1 = 1(1/4) = (cos2π + isin2π)1/4 = cos(π/2)+ isin(π/2) = i.

Therefore, squaring both sides it follows 1 = −1 as in the previous problem. Whatdoes this tell you about De Moivre’s theorem? Is there a profound difference betweenraising numbers to integer powers and raising numbers to non integer powers?

22. Review Problem 10 at this point. Now here is another question: If n is an integer, isit always true that (cosθ − isinθ)n = cos(nθ)− isin(nθ)? Explain.

23. Suppose you have any polynomial in cosθ and sinθ . By this I mean an expressionof the form ∑

mα=0 ∑

nβ=0 aαβ cosα θ sinβ

θ where aαβ ∈C. Can this always be writtenin the form ∑

m+nγ=−(n+m)

bγ cosγθ +∑n+mτ=−(n+m)

cτ sinτθ? Explain.

24. Suppose p(x) = anxn +an−1xn−1 + · · ·+a1x+a0 is a polynomial and it has n zeros,

z1,z2, · · · ,zn

listed according to multiplicity. (z is a root of multiplicity m if the polynomial f (x) =(x− z)m divides p(x) but (x− z) f (x) does not.) Show that

p(x) = an (x− z1)(x− z2) · · ·(x− zn) .

25. Give the solutions to the following quadratic equations having real coefficients.

(a) x2−2x+2 = 0

(b) 3x2 + x+3 = 0

(c) x2−6x+13 = 0