1wat bout cases like(-2)3/2
28 Answers
62@darkknight
if you have to find (-1)1/2 it is not real
while if you have to find (-1)2/7 it is real
also (-1)p/q depends a lot on whether P is even or odd...
1(-2)3/2 is clearly a complex no.
at all points where d simple-rational power has even no. in the denominator , d graph is discontinuous...!!!
isko plot karen 
62yes you are right..
complex numbers were not created to avaoid confusionbut to solve equations like
x2+1=0
:D
3COMPLEX NUMBERS
1.1 INTRODUCTION
We started our study of number systems with the set of
natural numbers, then the number zero was included to form
the system of whole numbers; negative numbers were defined.
Thus, we extended our number system to whole numbers and
integers.
To solve the problems of the type a ÷ b we included
rational numbers and consequently for some other problems
we defined irrational numbers. All the numbers taken together
are termed as real numbers. Also, these numbers can be
represented on a number line.
Now let us consider a situation where we want to solve
the equation.
x²+1 = 0
⇒
x²= −1
or
x= ± −1
In real numbers there is no solution to this and many
other such problems like the one given below
21ap/q = (ap)1/q = (a1/q)p
-12/2 = (-11/2)2
i2 = -1
but if √-12 = 1
62This is a very good doubt to have...
I got this doubt myself a lot of times....
Post7 is the best explanation that i was able to get..
3sir is the relation is only for real nos giving real values.
and also i2≠√-1x-1
3any no can be represented on the argand plane.
so in the argand the no -1 can be written as z=cos∩+isin∩.
and also z2/3 is wat i have written.First of all i want to know when did u got thiss doubt while revising complex nos or others
62yes this is a bit tricky..
think of this like this...
ap/q = (ap)1/q = (a1/q)p
now you will realise that the graph is defined for a lot of -ve values..
all irrational values it is not defined!
1it can be writen as \sqrt[3]{(-1)^{2}}
then 1 hi to hoga [7][7]
1Sir but the graph shows fractional power of -ve number is not defined?