Answer to Question #228328 in Complex Analysis for Leu

Question

Find all the cube roots of

Answer 1

Find all the cube roots of $1+i.$

The polar form of $1+i$ is $\sqrt{2}(\cos\dfrac{\pi}{4}+i\sin \dfrac{\pi}{4}).$

According to the De Moivre's Formula, all $n$-th roots of a complex number $r(\cos(\theta)+i\sin(\theta))$ are given by $r^{1/n}(\cos\dfrac{\theta+2\pi k}{n}+i\sin \dfrac{\theta+2\pi k}{n}), k=0, 1,...,n-1.$

We have that $r=\sqrt{2}, \theta=\dfrac{\pi}{4}, n=3.$

$k=0$

$$2^{1/6}(\cos\dfrac{\pi }{12}+i\sin\dfrac{\pi }{12})$$

$k=1$

$$2^{1/6}(\cos(\dfrac{\pi }{12}+\dfrac{2\pi}{3})+i\sin(\dfrac{\pi }{12}+\dfrac{2\pi}{3}))$$

$$=2^{1/6}(\cos(\dfrac{3\pi }{4})+i\sin(\dfrac{3\pi }{4}))$$

$k=2$

$$2^{1/6}(\cos(\dfrac{\pi }{12}+\dfrac{4\pi}{3})+i\sin(\dfrac{\pi }{12}+\dfrac{4\pi}{3}))$$

$$=2^{1/6}(\cos(\dfrac{17\pi }{12})+i\sin(\dfrac{17\pi }{12}))$$ $$=-2^{1/6}(\cos(\dfrac{5\pi }{12})+i\sin(\dfrac{5\pi }{12}))$$