Solve $z^{10}=i$ for $i$ with restrictions

Question

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.

Answer 1

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

Answer 2

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 $1$.

Answer 3

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

$$z = \left(z^{10}\right)^{\frac{1}{10}} = \dots$$