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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, 4 months ago

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For complex numbers, the "identity": \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) does not work. Clarence Chew · 3 years, 4 months ago

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@Clarence Chew Then why do we write for eg. sqrt. -7=sqrt 7 *i Kushagra Sahni · 3 years, 4 months ago

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@Kushagra Sahni 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 · 3 years, 4 months ago

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