Finding and graphing the cube roots of $1 + i$

Question

The question asks to find the indicated roots and graph them in the complex plane. The cube roots of $1 + i$

Thus, my answer is: \begin{align*} \omega_0 & = 2^{1/6}[\cos(\pi/12)+i\sin(\pi/12)]\\ \omega_1 & = 2^{1/6}[\cos(7\pi/12)+i\sin(7\pi/12)]\\ \omega_2 & = 2^{1/6}[\cos(13\pi/12)+i\sin(13\pi/12)] \end{align*}

and the graph I choose is attached below However, it would tell me that the answer of $\omega_0$ is right, whilst the others are wrong and I am not sure if the graph chose is right or wrong

What is the correct graph for this solution

choose the right graph

Answer 1

Since $$1+i= \sqrt 2 e^{i\pi /4}= \sqrt 2 e^{i( 2\pi + \pi /4)}= \sqrt 2 e^{i(4\pi+\pi /4)}$$

Therefore

$$ x^3 = 1+i \implies x_1= 2^{1/6} e^{i\pi /12} ; x_2= 2^{1/6} e^{i 9\pi /12} ; x_3= 2^{1/6} e^{i 17\pi /12}$$

Answer 2

Wir like to solve $z^3=1+i$ for $z\in\mathbb C$. We can do it using the polar coordinates, so $$ z^3=1+i=\sqrt{2}e^{\frac14\pi i}=\sqrt{2}e^{\frac14\pi i+2k\pi i}\text{ for all }k\in\mathbb Z. $$ We get $$ z=2^{\frac16}e^{\frac1{12}\pi i+\frac23k\pi i}\text{ for all }k\in\mathbb Z. $$ Because the exponential function is periodic to $2\pi i$, we get \begin{align*} z_0=&2^{\frac16}e^{\frac1{12}\pi i}=2^{\frac16}\left(\cos\left(\frac1{12}\pi\right)+i\sin\left(\frac1{12}\pi\right)\right),\\ z_1=&2^{\frac16}e^{\frac1{12}\pi i+\frac23\pi i}=2^{\frac16}e^{\frac{3}{4}\pi i}=2^{\frac16}\left(\cos\left(\frac34\pi\right)+i\sin\left(\frac34\pi\right)\right),\\ z_2=&2^{\frac16}e^{\frac1{12}\pi i+\frac23\cdot 2\pi i}=2^{\frac16}e^{\frac{17}{12}\pi}=2^{\frac16}\left(\cos\left(\frac{17}{12}\pi\right)+i\sin\left(\frac{17}{12}\pi\right)\right). \end{align*}

Why are your $w_1$ and $w_2$ wrong? You can simply compute: $$ w_1^3 =\left(2^{\frac16}e^{\frac7{12}\pi i}\right)^3=\sqrt{2}e^{\frac{21}{12}\pi i}=\sqrt{2}e^{\frac74\pi i}=1-i $$ and $$ w_2^3=\left(2^{\frac16}e^{\frac{13}{12}\pi i}\right)^3=\sqrt{2}e^{\frac{39}{12}\pi i}=\sqrt{2}e^{\frac54\pi i}=-1-i. $$