1/i = -i

Richy

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!

Mojo_Jojo

This just got solved in chat.

So

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

Dunadan019

Your method does not work because the calculation rule of square roots only works for real non negative roots.

Hirocon

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.

In your specific example,

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

Richy

Thanks all three of you.

Hirocon, 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!

Smug Duckling

A slightly different approach:

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

ED!

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.