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