Reciprocal of i ? (1 Viewer)

blah

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In the cambridge textbook, theres an example that says the reciprocal of i is both i and -i.

I have tried multiplying i by i/i and -i/-i and only end up with just -i.
How the hell do you get i.

thnx
 

McLake

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Originally posted by blah
How the hell do you get i.

thnx
It sneeaky, but think about it:

i = sqrt(-1)
1/i = 1/sqrt(-1)
= sqrt (1/-1)
= sqrt (-1)
= i

Nice isn't it ...
 

blah

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haha thnx. That is SO sneaky. Did you see that staright away?

Cause they don't give you an explanation in the textbook, maybe its meant to be easy?
 

McLake

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Originally posted by blah
Did you see that staright away?
Actually, I saw it quite quickly (and surprosed myself ...) I just started to think about how i meant the sqrt of -1 and that you could sqrt 1 to get 1 and it just followed on from there ...
 

wogboy

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How is it possible that 1/i = i?

if 1/i = i,

then i^2 = 1, (making i = 1 or -1)

which cannot be true since i^2 = -1 always.

If however we square both sides of the equation we get:

(1/i)^2 = i^2

1/i^2 = i^2

1/-1 = -1

-1 = -1, which is quite ok, but we aren't really allowed to square both sides of an equation since by doing that, we can prove:

1 = -1.

Something makes me quite suspicious about using those index/square root laws over the complex field.

In polar form,

i = cos(pi/2) + isin(pi/2)

therefore,

1/i = 1/(cos(pi/2) + isin(pi/2))

= (cos(pi/2) - isin(pi/2))/(cos^2(pi/2) + sin^2(pi/2))

= cos(pi/2) - isin(pi/2)/1

= -i

it feels wrong that 1/i = i
 
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spice girl

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the reciprocal of any complex number with mudulus 1, is the conjugate of that number.

so1/i = -i

think about it in mod-arg form
1 = cis0
i = cis(pi/2)
1/i = cis(0 - pi/2) = cis(-pi/2) = -i
 

Constip8edSkunk

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Re: Re: Reciprocal of i ?

Originally posted by McLake


It sneeaky, but think about it:

i = sqrt(-1)
1/i = 1/sqrt(-1)
= sqrt (1/-1)
= sqrt (-1)
= i

Nice isn't it ...

hmmm are you sure the surd rules apply to complex numbers? I half remember hearing my teacher say that they dont. Coz if they do, then i^2=(sqrt(-1))^2=sqrt(-1)*sqrt(-1)=sqrt(-1*-1)=sqrt(1)=1 and i^2 =/= 1

i could be wrong though, but this just doesnt look right.
 

McLake

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Whoa, it looks like no-one agrees with me ...

I could be wrong, but my solution looks right to me. Noticv eI don't apply surd rules over the comple really, since I convert i to sqrt (-1).
 

Constip8edSkunk

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blah: which page is it on?

McLake: when you say 1/sqrt(-1)=sqrt(1/-1) arent you using the surd rule sqrt(a)/sqrt(b) = sqrt(a/b)?
 

Constip8edSkunk

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ah, ic

I think it's not saying that i and -i are the reciprocals of i, but that i and -i are reciprocals
 

maniacguy

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McLake, the problem with your solution is that it doesn't take into account the negative square root idea.

Here sqrt(-1) = -i only.

You can't always make the assumption that both roots are valid.
 

KeypadSDM

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The problem is that you used a bad rule:

Sqrt[a]/Sqrt = Sqrt[a/b]
or re-arrange Sqrt[a]Sqrt = Sqrt[ab]

And i would have to say that even you know that this rule is incorrect when used with complex numbers...

.'. i^-1 = -i

Note:

Assuming above rule correct;

.'. Sqrt[-a]Sqrt[-a] = Sqrt[-a * -a]
LHS = -a
RHS = a

.'. a = -a
.'. 2a = 0
But a can be nonzero, just sub in 3 or something...
.'. 2 = 0
.'. 1 = 0
.'. Maths breaks down at this unfortunate point....

And how the hell can a number have 2 inverses -- it's absurd, because let a^-1 = b, and a^-1 = c
but from one of the fundamental rules of mathematics b = c
.'. -i = i ..... not good...
 

turtle_2468

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I think the reason some books say that is due to the inherent arbitrary nature of the complex number system. Let me try to explain... suppose you let j=-i. Then do lots of complex no. questions with j and -j instead of -i and i respectively. You'll find that they all produce the same solutions virtually... basically there are two complex numbers with the property that a) both of them squared equal minus one and b) the sum of them is zero, but who's to say which one is "positive"?? Because the word positive loses its meaning in complex numbers... so -i is only defined as the number that when summed with i makes zero... hence the confusion.. I hope that made _some_ sense to someone.
 

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