Proof of multiplicative inverse of complex number

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SUMMARY

The discussion focuses on deriving the formula for the multiplicative inverse of a complex number, represented as z = a + bi. The inverse is expressed as z-1 = u + vi, where the conditions (au - bv) = 1 and (av + bu) = 0 must be satisfied. The solution provided by a teacher is z-1 = a/(a2 + b2) - bi/(a2 + b2). Participants discuss methods to solve the system of equations, including matrix representation and substitution techniques.

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  • Understanding of complex numbers and their representation.
  • Familiarity with systems of linear equations.
  • Knowledge of matrix operations and row reduction techniques.
  • Basic algebraic manipulation skills.
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  • Learn how to derive the multiplicative inverse of complex numbers using algebraic methods.
  • Study matrix representation of linear equations and row reduction techniques.
  • Explore the geometric interpretation of complex numbers and their inverses.
  • Investigate alternative methods for solving linear systems, such as substitution and elimination.
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Students studying complex numbers, mathematics educators, and anyone interested in understanding the properties and operations involving complex numbers.

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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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