Prove a multiplicative inverse exists (complex number)

[glauss]

Homework Statement

Find c and d when:
a, b E R and a, b !=0

1/(a+bi)=c+di

Additionally, prove that (a+bi) has a multiplicative inverse.

Relevant Equations

1/(a+bi)=c+di

I have as a solution for part one:

c=(a)/(a^2 + b^2)
d=(-b)/(a^2 + b^2)

Which matches with the solution manual.

The manual goes on to give the solution for part b:

(a+bi) * ( (a)/(a^2 + b^2) - ((b)/(a^2 + b^2))i ) = 1

I'd simply like to know where the 'i' at the end of the second expression in the left part comes from.

 

[fresh_42]

Homework Statement:: Find c and d when:
a, b E R and a, b !=0

1/(a+bi)=c+di

Additionally, prove that (a+bi) has a multiplicative inverse.
Relevant Equations:: 1/(a+bi)=c+di

I have as a solution for part one:

c=(a)/(a^2 + b^2)
d=(-b)/(a^2 + b^2)

Which matches with the solution manual.

The manual goes on to give the solution for part b:

(a+bi) * ( (a)/(a^2 + b^2) - ((b)/(a^2 + b^2))i ) = 1

I'd simply like to know where the 'i' at the end of the second expression in the left part comes from.

The last equation is to check whether your solution is actually one, i.e. if $(a+bi)(c+di)=1$ where you have substituted your solution for $c$ and $d$. The $i$ is the one at the $d$. We have a complex number as solution.

 

[glauss]

Thanks for the reply.
I thought that because the 'i' was factored out in the solution for 'd', it wouldn't be a factor of the solution of 'd' when multiplying it by its inverse.
However, your solution makes sense - because we re-included 'i' as a factor of 'b' in the first term (a+bi), we should re-include it in the inverse bit for 'd'?
Does thus sound right?

 

[fresh_42]

Thanks for the reply.
I thought that because the 'i' was factored out in the solution for 'd', it wouldn't be a factor of the solution of 'd' when multiplying it by its inverse.
However, your solution makes sense - because we re-included 'i' as a factor of 'b' in the first term (a+bi), we should re-include it in the inverse bit for 'd'?
Does thus sound right?

A bit confused if you ask me, but right. c and d are real numbers, but c+di is complex and we started with the setting that 1=(a+bi)(c+di). During the calculation you used only real numbers, namely those to find c and d. At the end, we have to compose them again to the complex number which we were really looking for: c+di.

It starts to confuse me, too,

 

[FactChecker]

However, your solution makes sense - because we re-included 'i' as a factor of 'b' in the first term (a+bi), we should re-include it in the inverse bit for 'd'?
Does thus sound right?

That sounds confusing to me. The 'i' was there to start with. You solved for c and d in c+di. So the 'i' was already there and you just substituted your results in for c and d.

 

[glauss]

That makes perfect sense, thank you.