Question

iota(i)= root-1 and (i) square=1, how can we prove by (i).(i)= (i) square

Answer 1

To be clear are you asking:
Given \[\iota(i) = \sqrt{-1}\] and \[i^2 = 1\] prove \[i^2 = i * i\]

Answer 2

the notation is confusing me a little...

Answer 3

i = root(-1)

so (i)*(i) = root(-1)*root(-1) = [root (-1)]^2
\[(\sqrt{-1})^{2} = -1\]

Answer 4

I think in your question it shud be (i) square= -1 and NOT (i)square = 1

Answer 5

yes... sorry sir...

Answer 6

so
(i)(i) = -1 = (i)^2

Answer 7

no problem...
here u just hv to prove LHS = RHS

by the way kpelletij, where r u frm, I'm frm Delhi....

Answer 8

sir i am frm delhi to!! living in gandhinagar..As my father is srvicing here..
hmm sir how root-1*root-1=1

Answer 9

using the formula\[\sqrt{a}*\sqrt{b}=\sqrt{a*b}\]

\[\sqrt{-1}*\sqrt{-1}= \sqrt{-1*-1}=\sqrt{1}=1\]

Answer 10

@Harkirat

sqrt(-1) * sqrt(-1) = i * i = i^2 = -1

Your formula is only valid for real non-negative a and b.

Avoid the problem by converting to i at first opportunity.

Answer 11

"how can we prove by (i).(i)= (i) square"

You don't have to prove it, the definition of i is that i^2 = -1.

Meaning it has square roots of plus and minus i, either of which squared = i^2.