2
$\begingroup$

I had the equation:

$( 1 + 2 i ) w^2 + 4 w - ( 1 - 2 i ) = 0$

To find $w$ I did the following:

$( 1 + 2 i ) w^2 + 4 w - ( 1 - 2 i ) = 0$

$[( 1 + 2 i ) w - 1 ][ w + ( 1 - 2 i )] = 0$

Which gives:

$ w = 1/(1 + 2 i) = (1/5) - (2/5) i$

and

$ w = -(1 - 2 i) $

These are correct but in the mark scheme it shows the unsimplified roots as

$ w = 1/(1 + 2 i)$ and $ w = -[5/(1 + 2 i)] $

where did this second unsimplified root come from?

Edited The second unsimplified root is $-[5/(1 + 2 i)]$ and not $-[5/(1 - 2 i)]$ . Sorry!

  • 1
    The quadratic equation!!! – 2017-02-28
  • 1
    It should be $$-(1-2i)=\frac{-(1-2i)(1+2i)}{1+2i}=-\frac{5}{1\color{red}{+}2i}$$ – 2017-02-28
  • 0
    So you actually do have the same two roots, as the previous comment proved. – 2017-03-01

2 Answers 2

1

So, from this step \begin{equation}( 1 + 2 i ) w^2 + 4 w - ( 1 - 2 i ) = 0.\end{equation}

Apply the quadratic formula

\begin{equation} \frac{-b \pm \sqrt{b^2-4ac}}{2a} \end{equation}

where $a=( 1 + 2 i ),b=4, c=-( 1 - 2 i )$. You should be good to go.

1

The second root in the mark scheme is not equal to your second root

$w=-\frac{5}{1-2i} = -\frac{5}{1-2i} \frac{1+2i}{1+2i} = \frac{-5-10i}{5} = -1 -2i$

whereas it should be $-1+2i$ as you state.