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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 ?

goblinsly
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  • Check out the following link to help format your question.

    http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference

     – Vincent Sep 02 '14 at 23:17
  • Note that $i$ is identified with the point $(0,1)$, so the angle is $\pi/2$, not $-\pi/2$. – André Nicolas Sep 02 '14 at 23:25
  • Always start these kinds of problems by drawing the unit circle. Where is i? Now, take 1/6 of that angle. That's one of the sixth roots, and the others are powers of it. – user4894 Sep 02 '14 at 23:57
  • Sorry -i. Same principle, just take all angles in the usual counterclockwise direction from the positive x-axis. – user4894 Sep 03 '14 at 03:37
  • I will try to repeat my question and be as clear as possible. 270 and -90 degress denote a same angle. Why then will we get wrong solutions to the equation if we use angle 270, and right solutions if we use -90 ? – goblinsly Sep 03 '14 at 18:33

2 Answers2

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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 $$z \in \{e^{i\pi(4k+1)/12}\}_{k=0}^5$$.

graydad
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  • This answer is almost right, but you have to change $2k \to 4k$ This is because, for each $k$ you're adding $\pi$ instead of $2\pi$. – antonimmo Sep 03 '14 at 00:14
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$$z^6=i \iff z^6=e^{i(\pi/2+2k\pi)} \quad ,$$ 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.

beep-boop
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