Finding the roots of $(1 + i)^{\frac{1}{4}}$ The professor says that the $n = 4$ roots of this are in the form: $\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})$, where $k = 0, 1, 2, 3$.
So to find $\theta$, we find the $r = \sqrt{(1)^2 + (1)^2} = \sqrt{2}$ since $Re(1+i) = 1$ and $Im(1+i) = 1$. So $\sqrt{2}\cos\theta = 1$ and $\sqrt{2}\sin\theta = 1$, so the angle $\theta$ is $\frac{\pi}{4}$.
However, if we do $k=0$, then we get that one of the roots is $1$, which is obviously not true since $1^4 \neq 1 + i$. The professor says that the solutions are: $k=1: \cos(\frac{9\pi}{16}) + i\sin(\frac{9\pi}{16})$, $k=2: \cos(\frac{17\pi}{16}) + i\sin(\frac{17\pi}{16})$, and $k=3: \cos(\frac{25\pi}{16}) + i\sin(\frac{25\pi}{16})$. I plugged these into WolfRamAlpha and rose them to the $4$th power, but none of them return the form $1+i$.
What is incorrect about these steps?
 A: The $n$ $n$th roots of a nonzero complex number $z=r(\cos \theta+i\sin\theta)$ are given by
$$
w_k = \sqrt[n]{r}\left[\cos\left(\frac{\theta+2k\pi}{n}\right)+i\sin\left(\frac{\theta+2k\pi}{n}\right)\right],\tag{1}
$$
where $k=0,1,2,\ldots,n-1$. 
Problem: Find the four fourth roots of $z=1+i$. 
Solution. We have, in this case, $r=\sqrt{2}$ and $\theta=\arg(z)=\pi/4$. Using $(1)$ with $n=4$, we get
$$
w_k = \sqrt[8]{2}\left[\cos\left(\frac{\pi/4+2k\pi}{4}\right)+i\sin\left(\frac{\pi/4+2k\pi}{4}\right)\right],\quad k=0,1,2,3.
$$
Thus, we have the following:


*

*$k=0,\quad w_0=\sqrt[8]{2}\left[\cos\frac{\pi}{16}+i\sin\frac{\pi}{16}\right]$

*$k=1,\quad w_1=\sqrt[8]{2}\left[\cos\frac{9\pi}{16}+i\sin\frac{9\pi}{16}\right]$

*$k=2,\quad w_2=\sqrt[8]{2}\left[\cos\frac{17\pi}{16}+i\sin\frac{17\pi}{16}\right]$

*$k=3,\quad w_3=\sqrt[8]{2}\left[\cos\frac{25\pi}{16}+i\sin\frac{25\pi}{16}\right]$

A: $1+i = \sqrt{2}\cdot {e^{i\pi/4}}= \sqrt{2}\cdot e^{i\pi/4+i2k\pi}\to \left(1+i\right)^{1/4}= 2^{1/8}\cdot e^{i(\pi/16+k\pi/2)}, k=0,1,2,3.$
A: From the classic for $a+bi$ can be seen in polar form as 
\begin{align}
a + b i = \sqrt{a^{2} + b^{2}} \, e^{i \tan^{-1}(b/a)}
\end{align}
for which
\begin{align}
1 + i = \sqrt{2} \, e^{i \tan^{-1}(1)} = \sqrt{2} \, e^{\pi i/4} = 2^{1/2} e^{\pi i/4 + 2n \pi i}
\end{align}
for $n$ being integer. Now 
\begin{align}
(1+i)^{1/4} = 2^{1/8} e^{\pi i/16 + n\pi i/2} = 2^{1/8} \left[ \cos\left(\frac{(8n+1)\pi }{16} \right) + \sin\left(\frac{(8n+1)\pi }{16} \right) \right]
\end{align}
where $n = 0,1,2,3$
