This isn't urgent, but I'd like explanations + working out so I can understand the question. THanks!

Factorise:

2z^2 + 2(z-2)z +3/2 - 6i

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- Feb 9th 2006, 08:27 PMphgaoFactorisation
This isn't urgent, but I'd like explanations + working out so I can understand the question. THanks!

Factorise:

2z^2 + 2(z-2)z +3/2 - 6i - Feb 10th 2006, 12:01 AMCaptainBlackQuote:

Originally Posted by**phgao**

or using the quadratic formula to find its roots $\displaystyle a$ and $\displaystyle b$ then the

factorisation is:

$\displaystyle (z-a)(z-b)$.

So lets simplify the equation a bit:

$\displaystyle 2z^2 + 2(z-2)z +3/2 - 6\bold{i}=4z^2-4z+(3/2 - 6\bold{i})$

The only problem will be finding the square root of the discriminant.

In this case the discriminant is:

$\displaystyle 16(6 \bold{i}-1/2)=8(12\bold{i}-1)$

We need the square root of this. We proceed by converting the discriminant

to polar form, then finding the square root by square-rooting the modulus and

halving the argument. Or as MathWorld would put it:

$\displaystyle \sqrt{x+iy}=$$\displaystyle \pm(x^2+y^2)^{1/4}(\cos(1/2\ \mbox{atan} (y/x))+\bold{i} \sin(1/2\ \mbox{atan} (y/x)))$,

Which should allow you to complete the factorisation.

RonL