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# 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 ?

Note by Kushagra Sahni
3 years, 9 months ago

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

- 3 years, 9 months ago

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

- 3 years, 9 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.

- 3 years, 9 months ago