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?

#Algebra #NumberTheory #ImaginaryNumbers #SquareRoot

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Comments

wait how is 1/i equal to negative i?

Jonathan Hsu - 6 years, 1 month ago

Multiple 1/i by i/i

Aryan Gaikwad - 5 years, 9 months ago

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 - 6 years, 1 month ago

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 - 6 years, 1 month ago

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

Otto Bretscher - 6 years, 1 month ago

root(a/b) = roota/rootb

only for positive numbers

Dev Sharma - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$