How can you locate the intricate origins of i?

A variant of the Root of Unity trouble.

I intend to locate all feasible response to this:

$$z^n = i$$

Where $$i^2 = -1$$

2019-05-13 03:17:35
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Answers: 3

Generally, the solutions would certainly be of the kind


where $\omega_n=\exp\left(\frac{2i\pi}{n}\right)$ is an origin of unity, and also $j=0\dots n-1$.

2019-05-17 15:24:34

Also, observe that if $z^n=i$ after that $z^{4n}=1$. Hence, the complex numbers you are seeking are certain $4n$ - th origins of $1$.

If you recognize that the $m$ - th origins of 1 (any kind of $m$) can be created as powers of a solitary well - picked one (a primitive origin), it should not be also hardto locate specifically which $4n$ - th origins have actually the wanted building.

2019-05-17 15:18:01

If the polar kind of $z$ is $$z=r(\cos\theta + i\sin\theta),$$ there are $n$ distinctive remedies to the formula $w^n = z$ : $$w=\sqrt[n]{r}(\cos\frac{\theta +2\pi k}{n}+ i \sin\frac{\theta +2\pi k}{n}),$$ where $k=0,1,...,n-1$. In your instance, $z=i$, whose polar kind is offered by $r=1$, $\theta = \pi /2$.

2019-05-17 13:57:44