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Richy
Registered User regular

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!

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!

0

## Posts

So

As i/i =1, you can say

1/i = 1/i * i/i

= i/(-1)

=-i

Mojo_JojoonDunadan019on1/(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.

In your specific example,

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

HirocononHirocon, your reply also answered another equivalence that was bugging me that I hadn't asked, that I was going to work on after lunch. Awesome!

Richyoni * i = -1 (by definition)

=> i = -1 / i

=> -i = 1 / i

Smug DucklingonIt 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.

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