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# 1/i = -i

Registered User regular
edited April 2011
Where is the complex number, the square root of -1.
i = SQRT(-1)

1/i = -i is correct, as far as I know. But I can't work out a demonstration for it. Worse, when I try, I get the wrong result and I can't figure out why:

1/i = 1/SQRT(-1) = SQRT(1)/SQRT(-1) = SQRT(1/-1) = SQRT(-1/1) = SQRT(-1)/SQRT(1) = i/1 = i

I'd appreciate it if someone could point out where I'm going wrong there. And also prove that 1/i = -i. Thanks!

Richy on

## Posts

• A lemon squeezed in the salty fist of Poseidon Registered User regular
edited April 2011
This just got solved in chat.

So

As i/i =1, you can say
1/i = 1/i * i/i
= i/(-1)
=-i

Mojo_Jojo on
Homogeneous distribution of your varieties of amuse-gueule
• Registered User regular
edited April 2011
Your method does not work because the calculation rule of square roots only works for real non negative roots.

• Registered User regular
edited April 2011
In general, for real numbers a and b where at least one of either a or b is nonzero,

1/(a + bi) = (a - bi)/(a^2 + b^2).

You can see that the above forumla works by multiplying both sides by (a + bi).

1 = (a + bi)(a - bi)/(a^2 + b^2) = (a^2 + b^2)/(a^2 + b^2) = 1.

1/i = 1/(0 + 1i) = (0 - 1i)/(0^2 + 1^2) = -i/1 = -i.

Hirocon on
• Registered User regular
edited April 2011
Thanks all three of you.

Richy on
• Registered User regular
edited April 2011
A slightly different approach:

i * i = -1 (by definition)
=> i = -1 / i
=> -i = 1 / i

Smug Duckling on
• Registered User regular
edited April 2011
Richy wrote:
1/i = -i is correct, as far as I know.

It is correct, because 1/i = (i)^-1. This means "the multiplicative inverse of 'i'" - so what number do you have to multiply "i" by to get 1, - i.
Hirocon wrote: »
In general, for real numbers a and b where at least one of either a or b is nonzero,

1/(a + bi) = (a - bi)/(a^2 + b^2).

A nitpick (especially if this is asked on an exam): you would not show that 1/z = z/(a^2+b^2) in the above manner, since you are assuming what you are trying to show in the first place. Both sides are of course going to be equal since you already assumed they were equal to begin with: so multiplying 1/z by z to get 1 on the LHS will trivially give you 1 on the RHS.

ED! on
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