[B]complex numbers problem. PLEASE HELP

nairdarakiram

Hi I really need help with this maths problem ASAP.

Please tell me how to do it.

Given that á=1+3i is a root of the equation z²-(p+2i)z+q(1+i)=0,
And that p and q are real, find p and q and the other root of the equation.


Please someone help me me on this one. Thanx

sikes

complex roots come in conjugate pairs. therefore the roots are

1+3i
1-3i

substitute one in for z and let real = real and im = im

3=p
5=q

(i think)

Celtic Tiger

Complex numbers only occur in conjugate pairs when coefficients are real (in this case they aren't) so just sub in 1+3i and same method as above.

nairdarakiram

Thanx lads. I really needed the help. You were quick too. Thanx again.

sikes

ah neine. sorry bout that mis-infomation about the conjugate pairs stuff. gotta go over that again!!

hussey

if you have not got the answer here it is :

substiute 1+3i for Z

so you get
(1+3i)(1+3i) - (p+2i)(1+3i) + q(1+i) = 0

this gives you

(-8+6i) - (p+3pi +2i -6) + (q+qi) = 0

gives gives you two Equations

-8 - p + 6 + q = 0 ... real = real ... q - p = 2
6 -3p - 2 + q = 0 ... img = img .... q - 3p = -4

now from this p = 3 and q = 5

now an equation is defined as
z^2 - (sum of roots)z + (product of roots) = 0

let teh second root be a+bi

so therefore

1+3i + a + bi = 3 + 2i

hence
a = 2, b = -1

the second root is (2 - i)