Question

i^2=-1
prove it

Answer 1

take the square root of both sides, and you have an Identity

Answer 2

by definition \[i =\sqrt{-1}\]

Answer 3

but see it
i^2=i*i=(sqrt -1)*(sqrt -1)=sq rt (-1*-1)=sq rt 1=1

Answer 4

how it can be -1???????

Answer 5

no my friend
\[x^a * x^b = x^{a+b}\]

Answer 6

\[i^2=i\times i=\sqrt{ -1}\times\sqrt{ -1}\neq\sqrt {-1\times-1}=\sqrt1=1\]

Answer 7

do you understand what is wrong with what you did ?

Answer 8

your concept is true but i think this law is holds only when x is real no. is'nt it?

Answer 9

sqrts dont join up so easily with complex numbers

Answer 10

why?
can u tell me its reason

Answer 11

can u?

Answer 12

i^2 = -1 is a definition, you don't have to prove it...

Answer 13

\[\sqrt{a+ib}\cdot\sqrt{\frac1{a+ib}}=\begin{cases} -1&\arg(a+bi)≥π\\ \text{undefined} & a=b=0\\1&\text{otherwise}\end{cases}\]

Answer 14

@UnkleRhaukus You have gone all yellow, I didn't recognize u:-