Author Topic: Quick complex factorising (Read 734 times) Tweet Share

kettles · « **on:** March 06, 2012, 08:37:42 pm »

Hey guys how would you factorise z^6-1 over C

b^3 · « **Reply #1 on:** March 06, 2012, 09:06:45 pm »

Firstly break it up using DOPS, then you have two linear factors, and two quadratic factors, complete the square for the quadratic factors, now to be able to do DOPS again on those terms, you need a minus sign in the middle, now remember that $i^{2} =-1$ . so that $\frac{3}{4}$ is the same as $-\frac{3}{4}i^{2}$ , now you can factorise the those terms.

You should get something like below.
$\begin{aligned}(z^{6}-1) & =(z^{3})^{2}-(1)^{2} \\ & =(z^{3}-1)(z^{3}+1) \\ & =(z-1)(z^{2}+z+1)(z+1)(z^{2}-z+1) \\ & =(z-1)(z+1)(z^{2}+z+(\frac{1}{2})^{2}-(\frac{1}{2})^{2}+1)(z^{2}-z+(\frac{1}{2})^{2}-(\frac{1}{2})^{2}+1) \\ & =(z-1)(z+1)[(z+\frac{1}{2})^{2}+\frac{3}{4}][(z-\frac{1}{2})^{2}+\frac{3}{4}] \\ & =(z-1)(z+1)[(z+\frac{1}{2})^{2}-\frac{3}{4}i^{2}][(z-\frac{1}{2})^{2}-\frac{3}{4}i^{2}] \\ & =(z-1)(z+1)(z+\frac{1}{2}-\frac{\sqrt{3}}{2}i)(z+\frac{1}{2}+\frac{\sqrt{3}}{2}i)(z-\frac{1}{2}-\frac{\sqrt{3}}{2}i)(z-\frac{1}{2}+\frac{\sqrt{3}}{2}i) \\ & =(z-1)(z+1)(z+\frac{1-\sqrt{3}}{2}i)(z+\frac{1+\sqrt{3}}{2}i)(z-\frac{1+\sqrt{3}}{2}i)(z-\frac{1-\sqrt{3}}{2}i) \end{aligned}$

Hope that helps

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Author Topic: Quick complex factorising (Read 734 times) Tweet Share

kettles

Quick complex factorising

b^3

Re: Quick complex factorising

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