Finding argument of power of a complex number

Question

If $z=\sqrt 3-i$ then I have to write $z^7$ in $a+ib$ form.
$z^7=2^7(cos({-7\pi\over 6})+i sin({-7\pi\over 6}))$.
What I don't understand is writing the argument of this. It should be in $-\pi<Argz\leq \pi$ ?
Then it can be written as $z^7=2^7(cos(-(\pi+{\pi\over 6}))+i sin(-(\pi+{\pi\over 6})))=2^7(-cos({\pi\over6})+isin({\pi\over6}))$.
Or since ${-7\pi\over 6}$ is measured in the clockwise direction and falls in the second quadrant like this, should it be $z^7=2^7(cos({5\pi\over 6})+i sin({5\pi\over 6}))$.
Can someone please let me know how to find the argument in these cases?

Answer 1

Hint:

$$\dfrac{5\pi}6-\left(\dfrac{-7\pi}6\right)=2\pi$$

We can choose any argument of the form $2m\pi+\dfrac{5\pi}6$ where $m$ is any integer

as $\cos\left(2m\pi+\dfrac{5\pi}6\right)+i\sin\left(2m\pi+\dfrac{5\pi}6\right)=\cos\dfrac{5\pi}6+i\sin\dfrac{5\pi}6$

Answer 2

Let us first obtain the answer by direct computation (squarings, then division):

$$z=\sqrt3-i\to z^2=2-2\sqrt3 i\to z^4=-8-8\sqrt3 i\to z^8=-128+128\sqrt3 i\to \\z^7=-64\sqrt3+64i.$$

Now by the polar form,

$$z=2e^{-i\pi/6}$$

and

$$z^7=2^7e^{-i7\pi/6}=128\left(\cos\frac{7\pi}6-i\sin\frac{7\pi}6\right)=128\left(-\cos\frac{\pi}6+i\sin\frac{\pi}6\right) \\=-64\sqrt3+64i.$$

Note that to compute the power, the argument need not be normalized to the principal value as you take the sine/cosine.

$$\cos\frac{7\pi}6-i\sin\frac{7\pi}6=\cos\left(-\frac{5\pi}6\right)-i\sin\left(-\frac{\pi}6\right).$$