The Student Room Group

Why does |i|=1?

Can anyone explain to me why i=1|i|=1? I understand that x=x|x|=x if x>0x>0 and x=x|x|=-x if x<0x<0, but ii is neither positive or negative, so how can we find the absolute value of it?
Original post by A Confused Guy
Can anyone explain to me why i=1|i|=1? I understand that x=x|x|=x if x>0x>0 and x=x|x|=-x if x<0x<0, but ii is neither positive or negative, so how can we find the absolute value of it?


The modulus function, x=x|x|=x if x>0x>0 and x=x|x|=-x if x<0x<0, is correct as you defined it when x is a real number. It is defined differently for complex numbers: namely, for the complex number a+bi, it is defined as a2+b2\sqrt{a^2+b^2} (the distance between it and the origin on the Argand diagram. So from this, clearly |i|=1
a+bi=a2+b2|a+bi|=\sqrt{a^2+b^2}
i=0+1i=12=1|i|=|0+1i|=\sqrt{1^2}=1

Like omghacklol says, what you have wrote is only valid for real numbers. :smile:
(edited 8 years ago)
The other two answers may be considered slightly more "mathematical" but if you consider the modulus function simply to find the magnitude of a value, then you know the value of i = root(-1) but since we're just looking at the magnitude, or amount, then this is equal to root(1) which is just 1.
Original post by A Confused Guy
Can anyone explain to me why i=1|i|=1? I understand that x=x|x|=x if x>0x>0 and x=x|x|=-x if x<0x<0, but ii is neither positive or negative, so how can we find the absolute value of it?

The notion generalises massively. Don't expect to understand the following, but you might possibly have seen some of the words before. I'm just giving it to show that the absolute value is not an arbitrary choice of definition; it is the reason why z2=zzˉ|z|^2 = z \bar{z}. EDIT: if it was unclear, the fact that i2=1|i|^2 = 1 is an instance of the more general rule that z2=zzˉ|z|^2 = z \bar{z}, where zˉ\bar{z} is the complex conjugate.

The norm of a number in field K(θ)K(\theta) is the product of all the ϕi(θ)\phi_i (\theta) where ϕi\phi_i ranges over the embeddings (that is, KK-homomorphisms) K(θ)CK(\theta) \to \mathbb{C}. For K=QK = \mathbb{Q}, θ=1\theta = \sqrt{-1}, there are two C\mathbb{C}-embeddings (given by 1±i\sqrt{-1} \mapsto \pm i), so the norm of 1\sqrt{-1} is i×i=1-i \times i = 1.
(edited 8 years ago)
if you think of the modulus as the distance from the origin you will not go far wrong
Okay, that's really helped. Thanks guys. One other question though: why is 1i=i\frac{1}{i}=-i?


Posted from TSR Mobile
Original post by A Confused Guy
Okay, that's really helped. Thanks guys. One other question though: why is 1i=i\frac{1}{i}=-i?


Posted from TSR Mobile


1i=1iii=i1=i\displaystyle \frac{1}{i}= \frac{1}{i} \cdot \frac{i}{i}= \frac{i}{-1}=-i
Original post by A Confused Guy
Okay, that's really helped. Thanks guys. One other question though: why is 1i=i\frac{1}{i}=-i?


1i=i2i\displaystyle \frac{1}{i}=\frac{-i^2}{i}
Original post by Mr M
1i=i2i\displaystyle \frac{1}{i}=\frac{-i^2}{i}


You've got a typo in there - 1i=i2i=1i\frac{1}{i}=\frac{i^2}{i}=\frac{-1}{i}
Original post by BlueSam3
You've got a typo in there - 1i=i2i=1i\frac{1}{i}=\frac{i^2}{i}=\frac{-1}{i}


Nope.
Reply 11
Original post by BlueSam3
You've got a typo in there - 1i=i2i=1i\frac{1}{i}=\frac{i^2}{i}=\frac{-1}{i}


You appear to have 'proved' that 1i=1i\frac{1}{i}=\frac{-1}{i} and hence that 1=11 = -1 :smile:
Original post by davros
You appear to have 'proved' that 1i=1i\frac{1}{i}=\frac{-1}{i} and hence that 1=11 = -1 :smile:


I should not post after scout camps, sorry.

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