Proof of multiplicative inverse of complex number

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Homework Help Overview

The discussion revolves around deriving the formula for the multiplicative inverse of a complex number, represented as z = a + bi. The original poster expresses confusion about how to derive the inverse from the equations resulting from the product of z and its inverse equating to 1.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting up a system of linear equations based on the conditions for the multiplicative inverse. Some suggest using matrix methods, while others propose more direct algebraic manipulations.

Discussion Status

Some participants have provided guidance on alternative methods to approach the problem, including using linear algebra and matrix representation. The original poster has acknowledged a successful resolution using one suggested method but continues to seek clarity on the matrix approach.

Contextual Notes

There is an indication of confusion regarding the setup of the matrix for the linear equations derived from the conditions of the multiplicative inverse, as well as a mention of "major brain fog" affecting the original poster's ability to proceed.

philnow
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Homework Statement


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
Don't know how to get the last part... major brain fog. Please help?
 
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It's just a system 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.
 
philnow said:

Homework Statement


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
Don't know how to get the last part... major brain fog. Please help?

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.
 
Thanks Mark, I got the answer using your method :)

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:
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 getav + 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:
You're a champ, thanks a lot!
 

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