Solving a complex equation z^6 =- i

Question

I am doing some simple complex equations. z^6 = -i. Allright so i cant get angle from tan of angle = y / x because x = 0.

So i can get it either from x = z * cos of angle or y = z * sin of angle. First gives me 90 degress, second one -90. Second one is right since x = 0, y = -1.

But now here is something i dont understand. -90 (-Pi/2) is same as 270 (3Pi/2). But when i am solving this equation, if i use 270 instead of -90 i will get the wrong coordinates. Lets look at first coordinate when k = 0. Then angle is simply -Pi/12 in first case and Pi/4 in second case ( since its z^6 i had to divide all angles by 6). Well now from those 2 angles i dont get same coordinates anymore. This means that i HAVE to use -90 ( -Pi/2 ) not 270 ( 3Pi/2) when solving this equation. But how do i know that ?

Answer 1

Using Euler's equation we know $e^{i\pi(4k+1)/2}=\cos\left(\frac{\pi(4k+1)}{2}\right)+i\sin\left(\frac{\pi(4k+1)}{2}\right)=0+i*1=i.$ Thus $$z^{6}=e^{i\pi(4k+1)/2}$$ which means if we take the sixth root of each side we have $$z=e^{i\pi(4k+1)/12}=\cos\left(\frac{\pi(4k+1)}{12}\right)+i \sin \left(\frac{\pi(4k+1)}{12}\right)$$. You will get solutions for $k\in \{0,1,2,3,4,5\}$ after which point we can tell by looking at the trig functions $k=6$ is the same solution as $k=0$ and $k=1$ is the same solution as $k=7$, and... $k=n$ is the same solution as $k=n+6$ in general. Your final answer can be given as something like $$z \in \{e^{i\pi(4k+1)/12}\}_{k=0}^5$$.

Answer 2

$$z^6=i \iff z^6=e^{i(\pi/2+2k\pi)} \quad ,$$ where $k \in \mathbb{Z}.$

Taking 6th roots of both sides gives us $$z=e^{i\left[\frac{(2k+1)\pi}{12}\right]} \tag{I}.$$

Now take any six consecutive integers ( e.g. $k \in \{0,1,2,3,4,5\}$ ), plug in each value of $k$ into $(\rm{I})$ and then use the identity $$e^{i(\theta)} \equiv \cos(\theta)+i\sin(\theta)$$ and you're done.