# Complex number

1. Jul 20, 2013

### ranjitnepal

please tell me the solution i did is right or wrong and why?

we know,
I=√-1
I2=√-1*√-1 =√(-1)2 = √1 = 1

2. Jul 20, 2013

### micromass

Last edited by a moderator: May 6, 2017
3. Jul 20, 2013

### lurflurf

Wrong, you cannot move the exponent inside the radical. That you arrived at -1=1 is a sign of a mistake.

Last edited: Jul 20, 2013
4. Jul 20, 2013

### micromass

A mistake? For sure. But a very interesting mistake. "Paradoxes" of these kind indicate that something very interesting is going on with complex exponents. In particular, they indicate that complex exponents (such as roots) are multivalued. Once you take this approach, all paradoxes vanish :tongue2:

5. Jul 20, 2013

### symbolipoint

Would it be enough just to accept the definition so that we use both of these:

i*i=-1 and i=sqrt(-1)

This is what happens with (-i).
(-i)*(-i)=(-1)(-1)*i*i=(1)*(-1)=-1
What went wrong there?
Apparantly i*i=-1 and (-i)(-i)=-1.

Still seem good. i*i still -1 and (-1)(-1)=1

Last edited: Jul 20, 2013
6. Jul 20, 2013

### hilbert2

Yeah, similarly, by naively taking the natural logarithm of both sides of equation $e^{0}=e^{2\pi i}$ one gets the result $0=2\pi i$.

7. Jul 20, 2013

### HallsofIvy

Well, there are your first two errors! To begin with, it is "i", not "I"!
(I suspect your editor automatically changed your "i" to "I". So did mine- I had to "fool" it by typing "ia", then going back and deleting the "a"!)
More importantly, it is a mistake to write "$a= \sqrt{-1}$ because there is no such number before we define "i" and you can't define a new number to be something that doesn't exist to begin with! Defining "i" to be "the number whose square is -1" is better because -1, at least, does exist before we define the complex numbers. But has the difficulty that once we start working with the complex numbers we find that every number, except 0, has two square roots and this does not tell us which of the two roots of -1 "i" is.

Better is to define the complex numbers to be the set of orderd pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ b, c+ d) and multiplication by (a, b)*(c, d)= (ac- bc, ad+ bc). We can then identify the real numbers with pairs of the form (x, 0) and "i" with (0, 1).

With the "ordered pairs" definition, above, [tex]i^2= (0, 1)(0, 1)= (0(0)-(1)(1), 0(1)+ 0(1))= (-1, 0)[tex] which we have already identified with the real number -1.