Question

I am asking a question about "Complex Number"
how this equation comes up:
i^2 = (square root of -1)^2 = square root of (-1)^2 = square root of 1 = 1
^ represents to the power.
Anyone please answer.

Answer 1

i^2 = -1

i = sqrt(-1)

Answer 2

i = sqrt(-1)

Answer 3

i = sqrt(-1)

i^2 = [sqrt(-1)]^2

i^2 = -1

Answer 4

[sqrt(-1)]^2 = -1^(2/2) = -1^1 = -1

Answer 5

\[i ^{2}=1\]
how is this possible?

Answer 6

\begin{array}c
i&i^2&i^3&i^4\\
i&-1&-i&1
\end{array}
and the powers of I just keep repeating thru

Answer 7

i^2 doesnt = 1

Answer 8

your downfall is in trying to define i as a variable; when i is already defined

Answer 9

go to this url : http://www.purplemath.com/modules/complex.htm
here the equation : \[i ^{2}=\left( \sqrt{-1} \right)^{2}=\sqrt{\left( -1^{} \right)^{2}}=\sqrt{1}=1\] is given. how is this coming???

Answer 10

they are not using "i" is a complex number; they are using it as a variable.

i^2 does not equal sqrt[(-1)^2]

Answer 11

i^2 isint 1...

Answer 12

to wit they expalin in the next part:

But this doesn't make any sense! You already have two numbers that square to 1; namely –1 and +1. And i already squares to –1. So it's not reasonable that i would also square to 1. This points out an important detail: When dealing with imaginaries, you gain something (the ability to deal with negatives inside square roots), but you also lose something (some of the flexibility and convenient rules you used to have when dealing with square roots). In particular, YOU MUST ALWAYS DO THE i-PART FIRST!

Answer 13

lol that was a very complicated way of saying that

Answer 14

:) thats the purple math for you :)

they give you a proof by absurdity

Answer 15

@amistre thank you.

Answer 16

hold up, i realized something though, check this out:

Answer 17

youre welcome

Answer 18

bye everyone

Answer 19

\[i ^{2} =(\sqrt{-1})^{2}= (\sqrt{-1})(\sqrt{-1}) = \sqrt{-1*(\sqrt{-1})^{2}}=\sqrt{-1*(repeat)}\]
at which point you would have to define that either \[(\sqrt{a})^{2}=a \] or \[(\sqrt{a})^{2}=\sqrt{a ^{2}}=\pm a\]

Answer 20

im guessing for imaginaries they defined the former rather than the latter

Answer 21

Do not agree with the 4th expression of the first line.\[(\sqrt{-1})(\sqrt{-1})\neq \sqrt{-1*(\sqrt{-1)^{2}}}\]