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for $z^2=-1$. we have $z^2=e^{i(\pi +2k\pi)}$=>$z=e^{i(\frac{\pi}{2}+k\pi)}=cos(\frac{\pi}{2}+k\pi))+i sin(\frac{\pi}{2}+k\pi))$,

we have here two roots . for $k =0$ we have $z=i$, And for $k=1, z=-i$

did I miss something here in terms of roots.?

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3 Answers

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Yes you missed something: the simplest way to solve this equation. Recall that if two complex numbers $a$ and $b$ satisfy $a^2=b^2$ then $a=b$ or $a=-b$. Now:$$z^2=-1=i^2\iff z=i\textbf{ or } z=-i$$ and it is over.

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$z=\pm i$ are all roots of this equation.

The fundamental theorem of algebra states that every single-variable polynomial of degree $n$ with complex coefficients has $n$ roots in $\mathbb{C}$.

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Further simplify your solution

$$z= \cos (\frac{\pi}{2}+k\pi) +i \sin(\frac{\pi}{2}+k\pi)= i\cos(k\pi)=\pm i$$

to see you have only two roots corresponding to even and odd $k$'s.

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