Further Maths A2

Ok I'm not sure if anybody else has done this/is doing this but I need a hand. Exercise 3A question 23 or 24 (can't remember which) says that alpha=-1+2j
show alpha^2 and alpha^3 and hence show that -1+2j is a root of the equation z^3+7z^2+15z+25=0
I could do this but the next question asks to show the 2 other roots of this equation. I put the equation into my calculator and the roots are (-1+2j),(-1-2j) and (-5)
I don't know how to get these though? What am I not doing??

Reply 1

DFranklin

If p(z) is a complex polynomial with real coefficients, then if p(a+ib) = 0, p(a-ib) = 0 as well. This is usually expressed as saying that roots occur in conjugate pairs.

So in this case since -1+2i is a root, so is -1-2i.

Now suppose

\alpha, \beta, \gamma

are the 3 roots of our polynomial. Then

(z-\alpha)(z-\beta)(z-\gamma) = z^3+7z^2+15z+25

. Comparing the constant term on both sides we find

-\alpha\beta\gamma = 25

.

In other words, the product of the 3 roots must be -25. But (-1+2i)(-1-2i) = 5. So the 3rd root must be -25/5 = -5.

Reply 2

bliek

damn thats impressive!

Reply 3

s_abbott

Oh right I see now. And I agree, very impressive indeed. I didn't even think of it like that at all. Well, I know where to post future questions now to make sure that I get a good answer lol