Values for $(1+i)^{2/3}$ This question might seem easier than I'm making it seem. But how many values are there for $(1+i)^{2/3}$? Do I let $z=(1+i)^{2/3}$ so that $z^3=2i$? I'm asked to write each in polar coordinates and in rectangular form and also sketch their location. Any help will be appreciated thank you.
 A: Here is how you advance
$$ z^{3} =(1+i)^{2}= 2i = 2e^{{i\pi/2}} = 2 e^{{i\pi/2}}  e^{i2k\pi} =2e^{i\pi/2+i2k\pi} $$
$$ \implies z = 2^{1/3}e^{\frac{i\pi/2+i2k\pi}{3}},\quad k = 0,1,2.  $$
A: $(1+i)$ on the complex plane is a vector of length $\sqrt{2}$ and angle $45^\circ$. When you take something to the power of $k$, you take its magnitude to that power and multiply its argument by $k$. Therefore, you get a vector of length $2^{1/3}$ and angle of $30^\circ$. For the other values, simply add $240^\circ$ to the argument.
A: For all $n\in \mathbb N, \rho,r\in [0, +\infty[$ and $\theta, \phi\in \mathbb R$ the following folds:
For all $n\in \mathbb N, \rho,r\in [0, +\infty[$ and $\theta, \phi\in \mathbb R$ the following folds:
$$\begin{align} \left(\rho e^{i\theta}\right)^n=re^{i\phi}&\iff \rho ^ne^{in\theta}=re^{i\phi}\\
&\iff \rho ^n=r\land \cos(n\theta)=\cos(\varphi) \land \sin(n\theta)=\sin(\phi)\\
&\iff \rho ^n=r\land \exists k\in \mathbb Z\left(n\theta=\phi+2k\pi\right).\end{align}$$
Thus, given $w=re^{i\phi}$, the set of its $n$ roots is $\left\{\root n\of r\exp\left(i\dfrac{\phi+2k\pi}{n}\right)\colon k\in \mathbb Z\right\}$ and equals $\left\{\root n\of r\exp\left(i\dfrac{\phi+2k\pi}{n}\right)\colon k\in \{0, \ldots, n-1\}\right\}$.
Now let $n=3, w=(1+i)^2$, write $w$ as $re^{i\phi}$ and use the above. Due to the periodicity of $z\mapsto e^{z}$, there will be three roots, which will form an equilateral triangle. See the sketch below.
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A: Yes, the values of $(1+i)^{2/3}$ as a multivalued expression are the solutions of $z^3=(1+i)^2=2i$. Now write $z=re^{i\varphi}$ in polar coordinates. By comparing absolute values you have
$$
r^3 = |z^3| = |2i| = 2,
$$
so $r=\sqrt[3]{2}$. Comparing arguments, you have
$$
3\varphi \equiv \arg(z^3) \equiv \arg(2i) \equiv \frac{\pi}{2}\mod 2\pi,
$$
which has $3$ solutions for $\varphi\in[0,2\pi)$.
