Basic complex factorisation

Question

Let's say I want to find all the roots of $f(z)=z^8-256$. Factorising it, I find $f(z)=(z-2)(z+2)(z^2+4)(z^4+16)$.

$z =\pm2,\,\pm2i$ is only 4 roots. Shouldn't there be another 4?

Answer 1

$$z^8=256=256\times1=256e^{i2k\pi}$$

$$z=(z^8)^{1/8}=(256e^{i2k\pi})^{1/8}=2e^{i2k\pi/8}$$

$$z=2e^{i\pi/8},2e^{2i\pi/8},2e^{4i\pi/8},2e^{6ik\pi/8},2e^{8i\pi/8},2e^{10i\pi/8},2e^{12i\pi/8},2e^{14i\pi/8}$$ Note that

$$e^{i\theta}=\cos\theta+i\sin\theta$$

So, we have

$$z=\pm2,\pm2i,\pm\sqrt2(1+i),\pm\sqrt2(1-i)$$

And $$f(z)=(z-2)(z+2)(z^2+4)(z^4+16)$$ Like you said $$z=\pm2,\pm2i$$ We have our $4$ more roots hidden in $z^4+16$ and those are $$z=\pm\sqrt2(1\pm i)$$

Answer 2

Starting where you left off, note that $$a^2 - b^2 = (a-b)(a+b)$$

continuing

$$\begin{align} f(z) &=(z-2)(z+2)(z^2+4)(z^4+16)\\ &=(z-2)(z+2)(z^2-4i^2)(z^4-16i^2) \\ &=(z-2)(z+2)(z-2i)(z+2i)(z^2-4i)(z^2 + 4i) \\ &=(z-2)(z+2)(z-2i)(z+2i)(z-2i^{1/2})(z+2i^{1/2})(z^2 - 4i^3) \\ &=(z-2)(z+2)(z-2i)(z+2i)(z-2i^{1/2})(z+2i^{1/2})(z - 2i^{3/2})(z + 2i^{3/2}) \\ \end{align}$$

Now calculating $i^{1/2}$ can be done multiple ways. Here are two:

• Solve $(a+bi)^2 = i$
• Write in polar form as $1 ~\angle~ \pi/2$ and use De Moivre's formula

Either way you should get $i^{1/2} = \sqrt 2 + i\sqrt 2 $ (and also $i^{1/2} = - \sqrt 2 - i \sqrt 2$, but that just gives the same set of roots in a different order).