ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: GodlyOne on September 27, 2017, 12:47:28 am

Title: Quick complex number question
Post by: GodlyOne on September 27, 2017, 12:47:28 am: Hi,

So if given the quadratic equation, ${z^2+4iz+5}$ the roots are 5i and -i however when factorising this equation over the complex field, I thought u put these roots into the assumed form of $z-a$ so it would be $(z-5i)(z+i)$ but the solution says the signs are switched which i am confused about because i thought linear complex factors can be formed by roots like usual 3u polynomials
Title: Re: Quick complex number question
Post by: jamonwindeyer on September 27, 2017, 07:05:52 am: Hey! I think you may have mixed up the signs of your roots?

$z=\frac{-4i\pm\sqrt{-16-20}}{2}=\frac{-4i\pm6i}{2i}=i, -5i$

It definitely does work just as you expect once the roots are correct ;D
Title: Re: Quick complex number question
Post by: RuiAce on September 27, 2017, 09:17:23 am: $\text{Try expanding out what you claimed the answer to be.}\\ \begin{align*}(z-5i)(z+i)&= z^2-5i+i-5i^2\\ &= z^2-4i+5\end{align*}$
$\text{There are ways of using the 2U/3U techniques}\\ \text{but you need to address the }i\text{ floating about}$
Title: Re: Quick complex number question
Post by: GodlyOne on September 27, 2017, 09:28:57 pm: Thanks, I just realised that I forgot about the Sum of roots being negative which is why i got the wrong signs.