Solve $z^{10}=i$ for $i$ with restrictions Solve $z^{10}=i$ whose argument is strictly between 120° and 180°.
I have no idea how to do these ones we were only taught how to evaluate.
 A: The solutions to $z^n=z_0$ can be represented on the complex-plane as an n-sided regular polygon centered at origin. The only thing we need to determine is the first vertex of the polygon.
So, let us try to find any one solution of $z^{10}=i$
$$z^{10}=i$$
$$z^{10}=e^{\frac{i\pi}{2}}$$
$$z=e^{\frac{i\pi}{20}}$$
So the required complex number is $z=e^{\frac{i\pi}{20}}$ which has a modulus of $1$ and argument of $ {\frac{\pi}{20}}$ radians. Plotting this on the complex-plane :

Now, we can complete the polygon as follows

As the required argument of the solution is between $120^0$ and $180^0$ it must lie in the green region, i.e. the complex number denoted by the green arrow

The argument of the complex number is perfectly calculated by @Parcly. This was just a visual/intuitive answer to your question
A: The modulus of $i$ is $1$, so the modulus of its ten roots will also be $1$.
The principal tenth root of $i$ has argument $90^\circ/10=9^\circ$. To get the other roots we add $360^\circ/10=36^\circ$ repeatedly to the argument, and only $36^\circ×4+9^\circ=153^\circ$ lies within the stated range.
Thus $z$ has argument $153^\circ$ and modulus $1$.
A: Hint
$$z^{10} = i$$
can be written as
$$z^{10} = e^{i\left(\frac{\pi}{2}+2\pi k\right)} \quad k \in \{0, \dots, 9\}$$
So what is
$$z = \left(z^{10}\right)^{\frac{1}{10}} = \dots$$
