more complex numbers (need help) (1 Viewer)

[blackfriday]

here's the questions of the week:

(i) a and b are real numbers such that the sum of the squares of the roots of the equation x^2 + (a+ib)x + 3i = 0 is 8. find all possiible pairs of a and b.

(ii) 1+ i is a root of the equation x^2 + (a+2i)x + (5+ib) = 0, where a and b are real. find the values of a and b.

i know im going to smack my head on the wall when someone points the obvious out to me. i feel bad because i got a test tomorrow, and the person who comes last gets kicked back to 3u.

 

[grimreaper]

here's the questions of the week:

(i) a and b are real numbers such that the sum of the squares of the roots of the equation x^2 + (a+ib)x + 3i = 0 is 8. find all possiible pairs of a and b.

(ii) 1+ i is a root of the equation x^2 + (a+2i)x + (5+ib) = 0, where a and b are real. find the values of a and b.

i know im going to smack my head on the wall when someone points the obvious out to me. i feel bad because i got a test tomorrow, and the person who comes last gets kicked back to 3u.

haha they cant do that (kick someone back to 3u).

for i)
say the roots are c and d
c^2 + d^2 = 8, cd=3i
solve these simultaneously to get values for the roots, from which you can work out a and b. But maybe theres a quicker way...

 

[grimreaper]

wait there is a another, maybe quicker, way:

(c + d)^2 = c^2 + d^2 + 2cd
you have values for c^2 + d^2 and cd so you can find (c + d)^2
Then you find the square root of that using the normal way and you should have values for a + ib

 

[Slidey]

i): let the roots be v and u

v^2+u^2=8
v+u=-a-ib
vu=3i

v=3i/u
-9/u^2+u^2=8
u^4-8u^2-9=0
(u^2+1)(u^2-9)=0
(u^2-i^2)(u+3)(u-3)=0
(u-i)(u+i)(u+3)(u-3)=0
u=+i, +3

vu=3i
u=i,
vi=3i, v=3
i.e. (u,v)=(i,3)

vu=3i
u=-i
-vi=3i, v=-3
i.e. (u,v)=(-i,-3)

vu=3i
u=3
3v=3i, v=i
i.e. (u,v)=(3,i)

vu=3i
u=-3
-3v=3i, v=-i
i.e. (u,v)=(-3,-i)

Now, u+v=-a-ib
we have for (u,v): (i,3), (-i,-3), (3,i), (-3,-i), so:
first point:
3+i=-a-ib, a=-3, b=-1, (a,b)=(-3,-1)
second:
-3-i=-a-ib, a=3, b=1, (a,b)=(3,1)
third:
3+i=-a-ib, a=-3, b=-1: (a,b)=(-3,-1)
fourth:
-3-i=-a-ib, a=3, b=1: (a,b)=(3,1)

2 of these four solutions for a and b are repeated, so (a,b)= (3,1) or (-3,-1)
i.e.: a=+3 and b=+1

 

[Slidey]

And I am stupid becuase I did it the long way. Short way:
v^2+u^2=8
v+u=-a-ib
vu=3i

v^2+u^2=(v+u)^2-2vu=(a+ib)^2-6i=8
(a+ib)^2=8+6i
a^2-b^2=8
2ab=6
ab=3

Solve simultaneously. Eh. That way is no quicker.

 

[blackfriday]

thanks heaps again

grimreaper: i hope you're right about not being able to be kicked back down.

 

[Slidey]

(ii) 1+ i is a root of the equation x^2 + (a+2i)x + (5+ib) = 0, where a and b are real. find the values of a and b

In this case just substitute x=1+i in.

(1+i)^2 + (a+2i)(1+i) + 5 + ib = 0
2i + a - 2 +2i + ai +5 + ib = 0
a+(a+b)i=-4i-3
a=-3
a+b=-4
-3+b=-4
b=-1
i.e.: (a,b)=(-3,-1)

 

[Slidey]

^ If you're in the course, which you obviously are, you can only be advised to move down. If they try to force you, you have avenues to take to stop it.