- #1

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we know,

I=√-1

I

^{2}=√-1*√-1 =√(-1)2 = √1 = 1

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- Thread starter ranjitnepal
- Start date

- #1

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- 0

we know,

I=√-1

I

- #2

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Please read this: https://www.physicsforums.com/showthread.php?t=637214 [Broken]

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- #3

lurflurf

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

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- #4

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

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

symbolipoint

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we know,

I=√-1

I^{2}=√-1*√-1 =√(-1)2 = √1 = 1

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

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- #6

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"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.

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

- #7

HallsofIvy

Science Advisor

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Well, there are your first two errors! To begin with, it is "i", not "I"!please tell me the solution i did is right or wrong and why?

we know,

I=√-1

(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 "[itex]a= \sqrt{-1}[/itex] because there is no such number before we define "i" and you

Better is to define the complex numbers to be the set of orderd

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.I^{2}=√-1*√-1 =√(-1)2 = √1 = 1

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