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Square Root of Imaginary Numbers (Proof Correction)

When I was researching on imaginary numbers, I came across this proof -

\(i=\sqrt { -1 } =\sqrt { -\frac { 1 }{ 1 } } =\frac { \sqrt { 1 } }{ \sqrt { -1 } } =\frac { 1 }{ i } =-i\)

which is clearly wrong. But even after analyzing it carefully, I couldn't find the error in it.

Where exactly is the error in the proof?

Note by Aryan Gaikwad
2 years, 7 months ago

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wait how is 1/i equal to negative i?

Jonathan Hsu - 2 years, 7 months ago

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Multiple 1/i by i/i

Aryan Gaikwad - 2 years, 3 months ago

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root(a/b) = roota/rootb

only for positive numbers

Dev Sharma - 2 years, 3 months ago

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For complex numbers (unlike non-negative real numbers) the square root is defined as a multi-values function: \(\sqrt{-1}=\pm{i}\). If \(z\) is any non-zero complex number, then \(\sqrt{z}\) has two values.

Does that help?

Otto Bretscher - 2 years, 7 months ago

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wait so if you take -1 and raise it to the 3/2 power, it would be i and -i? because -1^3/2 = (/sqrt{-1})^3/2 = /sqrt{-1} * /sqrt{-1} * /sqrt{-1} = -1 * /sqrt{-1} = -i, but it can't be i...

Jonathan Hsu - 2 years, 7 months ago

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Again, \((-1)^{3/2}=\pm{i}\) , with the "principal value" being \(-i\).

Otto Bretscher - 2 years, 7 months ago

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