Find an $v,w \in \mathbb{C}$ such that $(1+2i)v=1-i$ and $w^{2}-i=0$ Task from an old exam:

Find an $v,w \in \mathbb{C}$ such that $(1+2i)v=1-i$ and $w^{2}-i=0$
  and write the solutions in this form: $a+bi$

I'm currently trying to solve this, probably written several pages and always failed calculating it with success so I came here. I think there is some info missing for this task, about definition but anyway maybe you can give me some ideas of how this could be solved?


We obtain from $(1+2i)v=1-i$
\begin{align*}
v&=\frac{1-i}{1+2i}\\
&=\frac{(1-i)(1-2i)}{(1+2i)(1-2i)}\\
&=\frac{-1-3i}{5}\\
&=-\frac{1}{5}-\frac{3}{5}i
\end{align*}

We recall
\begin{align*}
w^2&=a+ib\\
&=r\left(\cos\phi+i\sin \phi\right)\\
w_{1,2}&=\pm\sqrt{r}\left(\cos\frac{\phi}{2}+i\sin \frac{\phi}{2}\right)\\
\end{align*}

We obtain from $w^2-i=0$
\begin{align*}
w^2&=i\\
&=1\left(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}\right)\\
\\
w_{1,2}&=\pm\sqrt{1}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)\\
&=\pm\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)
\end{align*}

