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Complex number $i=\sqrt{-1}$

Now i consider $$\frac{1}{i}=\frac{1}{\sqrt{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}=i$$

so $$i^2=1$$

Umesh shankar
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    This is due to the illness of the definition of $\sqrt{}$ sign over complex numbers. Generally, there are $n$ branches of the function $x^{\frac{1}{n}}$ – S. D. Z May 04 '17 at 02:41
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    The rule $\sqrt{u}/\sqrt{v}=\sqrt{u/v}$ does not generally hold for complex numbers $u$ and $v$. – anon May 04 '17 at 02:43
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    This post is good: https://www.reddit.com/r/askscience/comments/3k5125/why_exactly_is_1_1_11_1_1_i²_1_impossible/ – Jay Zha May 04 '17 at 02:45

1 Answers1

4

if $1=\sqrt{1}$ Then

$e^{i2\pi}=e^{i\pi}$

See your error?

The Dead Legend
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