SQUARE OF IOTA

Guys, we all know that square of i is obviously -1. But i= sqrt -1 so i^2 will be equal to sqrt -1 *sqrt -1 = sqrt 1=+1 or -1. And all books say i^2 is only -1. I wonder why. Anything wrong in this ?

Easy Math Editor

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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}`	$\boxed{123}$

Comments

For complex numbers, the "identity": $\sqrt{a}\times\sqrt{b}=\sqrt{ab}$ does not work.

Clarence Chew - 7 years, 3 months ago

Then why do we write for eg. sqrt. -7=sqrt 7 *i

Kushagra Sahni - 7 years, 3 months ago

Actually, we don't.

To get $\sqrt{-7}$ , we first say the answer is $ai+b$ .

Squaring $ai+b$ , we have $-a^2+b^2+2abi$ .

We can now compare complex parts. $ab=0$ . Hence $a=0$ or $b=0$ .

If a is 0, the result is positive. So the only way is b being 0.

Thus $a=\sqrt{7}$ , the negative rejected as it is not in the principal domain.

Clarence Chew - 7 years, 3 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}`	$\boxed{123}$