Question

question about complex numbers help pls

Answer 1

why is √ -5 = √ -1*5 = √ -1*√ 5 = i√ 5

Answer 2

pretty much by definition

Answer 3

isnt \[\sqrt{a}\sqrt{b}=\sqrt{ab} \] not defined for a or b<0 ?

Answer 4

\(i\) has no life of its own. it is another name for \(\sqrt{-1}\) with which is another way of saying \(i^2=-1\)

Answer 5

i dont understand why \[i \sqrt{5} = \sqrt{-1}\sqrt{5} = \sqrt{-1*5} = \sqrt{-5} \] if what i said above

Answer 6

\(\sqrt{a}\) is not a real number if \(a<0\) but if you are in the word of complex numbers, then you are allowing it

Answer 7

oh alright .. so only for complex

Answer 8

yes, you are dealing with complex numbers. so you are allowing the square root of negative numbers

Answer 9

why cant u do \[i^2 = \sqrt{-1}\sqrt{-1} = \sqrt{-1*-1} = \sqrt(1) = 1 ?\]

Answer 10

definition ?

Answer 11

notice that the statement
\[\sqrt{ab}=\sqrt{a}\sqrt{b}\] is false

Answer 12

it is true if \(a>0, b>0\) but not true if \(a<0,b<0\)

Answer 13

oh so what i just said for i^2

Answer 14

for example
\[\sqrt{36}=\sqrt{-6\times -6}\neq \sqrt{-6}\times \sqrt{-6}\]

Answer 15

yes exactly

Answer 16

oh alright

Answer 17

is \[\sqrt{ab}=\sqrt{a}\sqrt{b}\] true if one of a or b <0 ?

Answer 18

guys ??

Answer 19

when you multiply two complex number ... you add the argument and multiply the moduli

Answer 20

\[ \sqrt{36}=\sqrt{-6\times -6}\neq \sqrt{-6}\times \sqrt{-6} \]

this follows from this property
http://en.wikipedia.org/wiki/Exponentiation#Rational_exponents

Answer 21

and \( \sqrt{-6}\) <--- the moduli is \( \sqrt 6 \) and argument is \( \pi \over 2 \)

Answer 22

when you multiply
\[ \sqrt{-6} = \sqrt{6} e^{i {\pi \over 2}}\]
\[ \sqrt{-6} \times \sqrt{-6} = \sqrt{6} e^{i {\pi \over 2}} \times \sqrt{6} e^{i {\pi \over 2}} = 6 e^{i \pi}\]