What would be a formal way to prove the above statement?
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4What's your definition of $i$? – Botond May 12 '19 at 18:57
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3Also, what's your definition of $z^{\frac12}$? – Cameron Buie May 12 '19 at 18:59
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@Botond $i^2$ = -1 – bigsbylp May 12 '19 at 19:00
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You can just square both sides of the equality, although it's not a proof because $\left( (-1)^{\frac{1}{2}} \right)^2$ is also equal to $(-i)^2$. – Ertxiem - reinstate Monica May 12 '19 at 19:06
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$$\begin{align} (-1)^{1/2} &=e^{\ln{(-1)}/2}\\ &=e^{i\pi/2}\\ &=\cos{(\pi/2)}+i\sin{(\pi/2)}\\ &=i\\ \end{align}$$ Assuming the principal valued complex natural logarithm.
Peter Foreman
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