Proving geometric sum for complex numbers

[CGandC]

Homework Statement

Prove that for every complex number $z \neq 1$ it occurs that $\sum_{n=0}^{N} z^{n}=\frac{1-z^{N+1}}{1-z}$

Relevant Equations

$z= a+ib , z = r e^{i \theta}$

I went ahead and tried to prove by induction but I got stuck at the base case for $N =1$ ( in my course we don't define $0$ as natural so that's why I started from $N = 1$ ) which gives $\sum_{k=0}^1 z_k = 1 + z = 1+ a + ib$ .
I need to show that this is equal to $\frac{1- z^2 }{1-z}$ , i.e. $1+z = \frac{1- z^2 }{1-z}$.
So I went straight ahead and did as follows:
$\frac{ 1-(a+ib)^2 }{1-(a+ib) } = \frac{ 1-(a+ib)^2 }{1-a - ib) } \cdot \frac{1- a + ib}{ 1- a + ib } = \frac{ a^{3}+i a^{2} b-a^{2}+a b^{2}-2 i a b-a+i b^{3}+b^{2}+i b+1 }{a^{2}-2 a+b^{2}+1}$ but I don't really know how to continue from here.

I also tried using $z = r e^{i \theta }$:
$\frac{ 1 - r^2 e^{2 i \theta } }{ 1- r e^{i \theta } } = \frac{ 1 - r^2 e^{2 i \theta } }{ 1- r e^{i \theta } } \cdot \frac{ 1- r e^{-i \theta } }{1- r e^{-i \theta }} = \frac{r^{3} e^{i \theta}-r^{2} e^{2 i \theta}-r e^{-i \theta}+1}{r^{2}-2 r \cos (\theta)+1}$ and here I also stopped, unclear how to continue.

Can you please help? I don't know how to show the base case.

Edit: Problem's solved!

 

[vela]

Why don't you just factor the numerator?

 

[pasmith]

@CGandC: Complex numbers obey the usual rules of arithmetic. You don't need to split $z$ into real and imaginary parts. It is easier to expand $(1 - z)\sum_{n=0}^N z^n$ and show that this equals $1 - z^{N+1}$.

Why don't you just factor the numerator?

In my view that comes fairly close to assuming what the question asks you to prove.

 

[CGandC]

I factored and Indeed I got what I wanted:
$\frac{1- z^2 }{1-z} = \frac{ 1-(a+ib)^2 }{1-(a+ib) } = \frac{ 1-(a+ib)^2 }{1-a - ib) } \cdot \frac{1- a + ib}{ 1- a + ib } = \frac{ a^{3}+i a^{2} b-a^{2}+a b^{2}-2 i a b-a+i b^{3}+b^{2}+i b+1 }{a^{2}-2 a+b^{2}+1} = \frac{(a+1)\left(a^{2}-2 a+b^{2}+1\right)+i b\left(a^{2}-2 a+b^{2}+1\right)} {a^{2}-2 a+b^{2}+1} = a+1 + ib = 1 + a + ib = 1 + z$

But as @pasmith proposed, it would've been easier if I'd show from the start that $(1 - z)\sum_{n=0}^N z^n = 1 - z^{N+1}$
I haven't thought about that though.

that's it!, thanks for the help!

 

[Mark44]

I factored and Indeed I got what I wanted:
$\frac{1- z^2 }{1-z} = \frac{ 1-(a+ib)^2 }{1-(a+ib) } = \frac{ 1-(a+ib)^2 }{1-a - ib) } \cdot \frac{1- a + ib}{ 1- a + ib } = \frac{ a^{3}+i a^{2} b-a^{2}+a b^{2}-2 i a b-a+i b^{3}+b^{2}+i b+1 }{a^{2}-2 a+b^{2}+1} = \frac{(a+1)\left(a^{2}-2 a+b^{2}+1\right)+i b\left(a^{2}-2 a+b^{2}+1\right)} {a^{2}-2 a+b^{2}+1} = a+1 + ib = 1 + a + ib = 1 + z$

$1 - z^2 = (1 + z)(1 - z)$
I don't see that as assuming what the problem is asking you to prove. What you did is essentially the same as what I have above, although very much more long-winded.

 

[CGandC]

You are correct, it was a very silly mistake of mine ignoring that.. probably because I've sat through lots of math exercises today non-stop and learning new topics.

 

[vela]

In my view that comes fairly close to assuming what the question asks you to prove.

I was referring to factoring $1-z^2$ to verify the base case, not the general case.