# Proof of multiplicative inverse of complex number

1. Sep 2, 2009

### philnow

1. The problem statement, all variables and given/known data
I'm trying to derive the formula for the multiplicative inverse of a complex number. I say that:

z=a+bi
z-1=u+vi

and zz-1 must be 1 so

zz-1=(a+bi)(u+vi)
=(au-bv)+(av+bu)i
=1

So in order for zz-1=1, (au-bv) must be 1 and (av+bu) must be 0. My teacher has solved this as:

z-1= a/(a2+b2) - bi/(a2+b2)

I just don't know how to get that from au-bv=1 and av+bu=0

2. The attempt at a solution

2. Sep 2, 2009

### aPhilosopher

It's just a sytem of two linear equations in the two unknowns u and v. You can set it up as a matrix and do row reduction or you can do it the old fashioned way.

3. Sep 2, 2009

### Staff: Mentor

I don't see that you have used the fact that z-1 = 1/(a + bi). Multiply the latter expression by 1 in the form of (a - bi)/(a - bi). See if that helps.

4. Sep 2, 2009

### philnow

But, if I were to solve this using linear algebra, what would my matrix look like? I think:

a -b 1
b a 0

That doesn't look right, and I'm having difficulties solving this. Any pointers? Thanks in advance.

Last edited: Sep 2, 2009
5. Sep 2, 2009

### aPhilosopher

a matrix is overkill for this one.

you have au - bv = 1 and av + bu = 0

so u = (1 + bv)/a. substitute this into the second one to get

$$av + b\frac{(1 + bv)}{a} = 0$$

$$v(a + \frac{b^2}{a}) = -\frac{b}{a}$$

$$v\frac{a^2 + b^2}{a} = -\frac{b}{a}$$

you can probably take it from there.

If you do want to do it with matrices, the first step is to multiply the first row by -b/a and add it to the second row. That should get you going.

Last edited: Sep 2, 2009
6. Sep 2, 2009

### philnow

You're a champ, thanks a lot!