complex number questions (1 Viewer)

[relativity1]

The complex number z=1+2i is a root of the equation z^2-aiz+b=0 where a and b are real numbers.
i) find values of and b
ii) Find root of the equation

 

[calamebe]

I'll try and do it soon, but first of all you would substute z=1+2i into the equation, to get an equation for a and b.

 

[Ekman]

Sub the root in and you should get something like this:



And for this to be true:



And you can solve to get a=4, b=-5

For ii, you can use sum of roots:

 

[calamebe]

Ok so first of all substitute in z = 1 + 2i

(1+2i)^2 - ai(1+2i) + b = 0
1 + 4i -4 - ai +2a + b = 0
(2a + b - 3) + i(4 - a) = 0 (real and imaginary parts)
as both the real and imaginary parts have to be 0, we know:
2a + b - 3 = 0 and 4 - a = 0
So from the right equation, we get a = 4, so we substitute this value into the left equation:
8 + b - 3 = 0
b = -5

so a = 4 and b = -5

For part ii):

We now know the equation is z^2 - 4iz -5 = 0
So we can use the quadratic formula to find the roots:
z=(4i ± root(-16-4x-5))/2
z=(4i ± root(4))/2
z=(4i ± 2)/2
so z = 1 + 2i or -1 + 2i

 

[calamebe]

Sub the root in and you should get something like this:



And for this to be true:



And you can solve to get a=4, b=-5

For ii, you can use sum of roots:

Haha didn't even think of doing the sum of roots.