# Thread: Polynomial with complex roots

1. ## Polynomial with complex roots

Hey guys I'm not sure whether my approach to this question is right...kinda looks simple enough eh?

QUESTION
One root of $\displaystyle x^{2}+(1-i)x+K=0$ is $\displaystyle 2+3i$. Find the other root of this equation and the value of K in the form $\displaystyle a+ib$

So this is what I did:

Since $\displaystyle 2+3i$ is a root, $\displaystyle 2-3i$ is also a root as complex roots come in conjugate pairs

$\displaystyle (x-(2+3i))(x-(2-3i))=x^{2}-4x+13$

But then?
$\displaystyle \Sigma\alpha=\alpha+\beta=(2-3i)+(2+3i)=\frac{-b}{a}=4\neq1-i$

have I misunderstood something?...im rusty with complex numbers

The actual answers are that the other root is $\displaystyle -3-2i$,and $\displaystyle K=-13i$

Thanks

2. Originally Posted by aonin
Since $\displaystyle 2+3i$ is a root, $\displaystyle 2-3i$ is also a root as complex roots come in conjugate pairs

That is true if the polynomial has real coefficientes, an this is not the case. Substitute $\displaystyle x=2+3i$ and you will find $\displaystyle K$.

Fernando Revilla

3. If a polynomial has REAL coefficients, then what you said was correct. That is, roots come in complex conjugate pairs. However, this polynomial has a complex coefficient, so what you said doesn't hold (sorry!).

Personally, I would just work out $\displaystyle (x-2-3i)(x-a-bi)$, using the fact that the coefficient of $\displaystyle x$ must be $\displaystyle (1-i)$. It is, perhaps, crude. But it works.

4. Originally Posted by aonin
Hey guys I'm not sure whether my approach to this question is right...kinda looks simple enough eh?

QUESTION
One root of $\displaystyle x^{2}+(1-i)x+K=0$ is $\displaystyle 2+3i$. Find the other root of this equation and the value of K in the form $\displaystyle a+ib$

So this is what I did:

Since $\displaystyle 2+3i$ is a root, $\displaystyle 2-3i$ is also a root as complex roots come in conjugate pairs

$\displaystyle (x-(2+3i))(x-(2-3i))=x^{2}-4x+13$

But then?
$\displaystyle \Sigma\alpha=\alpha+\beta=(2-3i)+(2+3i)=\frac{-b}{a}=4\neq1-i$

have I misunderstood something?...im rusty with complex numbers

The actual answers are that the other root is $\displaystyle -3-2i$,and $\displaystyle K=-13i$

Thanks

Put 2+3i in the equation and find the value of k -> k=-3i

then put the k in the equation and find two X (quadratic function)
note that sqrt(50i)=5+5i

5. Originally Posted by ahaok
Put 2+3i in the equation and find the value of k -> k=-3i

then put the k in the equation and find two X (quadratic function)
note that sqrt(50i)=5+5i
Thanks!

6. Originally Posted by aonin
Hey, I got this..and I've not learnt how to obtain the roots so could you show me the working out after this step pls

$\displaystyle k=-13i$

$\displaystyle x^{2}+(1-i)x-13i=0$

Im not sure how to find the roots? uisng square roots of complex numbers lol? let $\displaystyle z^{2}=x^{2}-y^{2}+2xyi$?
infact you must ontain the roots of a quadratic function:
Quadratic function - Wikipedia, the free encyclopedia
in this case:
a=1, b=1-i, c=-13i

X1,2=(-b+-sqrt(b^2-4ac))/2a

7. Originally Posted by ahaok
infact you must ontain the roots of a quadratic function:
Quadratic function - Wikipedia, the free encyclopedia
in this case:
a=1, b=1-i, c=-13i

X1,2=(-b+-sqrt(b^2-4ac))/2a
This is all a bit too much though - as the OP pointed out, to do it this way you need to find the square root of a complex number, and so we are spiralling down a route that, although correct, requires a lot of work.

It is much easier to approach the question in the way I pointed out in my post.

$\displaystyle (x-2-3i)(x-a-bi) = x^2 + (1-i)x + K$
$\displaystyle \Rightarrow x^2 - (a+2+(b+3)i)x +(2+3i)(a+ib) = x^2 + (1-i)x + K$
$\displaystyle \Rightarrow -(a+2+(b+3)i) = 1-i$

So, what are a and b? So what is the other root, and what is K?

Simples...