Problem
Find the roots of $$z^3 = -1 - i$$
And calculate $$ \sqrt[3]{-1-i}$$
I'm looking at the solution outlined in my book but I'm having problems understanding it.
I can find the length and argument of $$-1-i$$ but then what? I guess the length is $$\sqrt{2}$$ and $$Arg(z^3) = \frac{3 \pi}{2}$$ My book says the argument is $$\frac{-3 \pi}{4}$$ though, not sure why.
On
If $r(\cos\theta+i\sin\theta)=-1-i$ where real $r>0$ and $\pi<\theta\le\pi$
we have $\displaystyle r\cos\theta=r\sin\theta=-1\implies r^2=(r\cos\theta)^2+(r\sin\theta)^2=(-1)^2+(-1)^2=2$
$\displaystyle\implies \cos\theta=\sin\theta=-\frac1{\sqrt2}<0$ so $\theta$ lies in the third Quadrant
Using this, $\theta=\arctan\left(\dfrac{-1}{-1}\right)-\pi=\dfrac\pi4-\pi$
Now use this
Let $z = re^{i \theta}$. Then it follows $r = (2)^{1/6}$. Divide out $r$ to get $e^{i 3\theta} = e^{-i \frac{3 \pi}4} $. (The angle follows from basic right triangle stuff). Now just set $3 \theta = -3 \pi/4$ and find the three angles from $0$ to $2 \pi$.