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An introduction to Complex numbers

May 3, 2020 Mouctar Algebra 0

Binomial theorem

In order to complete our complex numbers class, we will touch slightly the binomial theorem.
If we have to take $m$ elements $n$ at a time, we can write it the following way:

$\binom mn=^mC_n=\frac{m!}{n!(m-n)!}$

With $n$ and $m$ being integers and $0\leq n \leq m$

As example:
$\binom 42=\frac{4!}{2!(4-2)!}=6$

Please note the following rules:

$\binom n0=1$
$\binom nn=1$
$\binom n1=n$
$\binom n{n-1}=n$
$0!=1$

This can be used to generate what we call Pascal triangle.

The binomial theorem is shown as follows:

$(a+b)^n=\sum\limits_{k=0}^n \binom nk a^{n-k}b^k$

Example1:

$(a+b)^3$

Here we have $n=3$

$\binom 30=1$
$\binom 31=3$
$\binom 32=3$
$\binom 33=1$

$(a+b)^3=\binom 30 a^3b^0+\binom 31 a^2b^1+\binom 32 a^1b^2+\binom 33 a^0b^3$

Finally:

$(a+b)^3=a^3+3a^2b+3ab^2+b^3$

Example2:

Find $(x+y)^7$

$(x+y)^7=\binom 70 x^7y^0+\binom 71 x^6y^1+\binom 72 x^5y^2+\binom 73 x^4y^3+\binom 74 x^3y^4+\binom 75 x^2y^5+\binom 76 x^1y^6+\binom 77 x^0y^7$

$\binom 70=\frac{7!}{0!(7-0)!}=\frac{7!}{7!}=1$

$\binom 71=\frac{7!}{1!(7-1)!}=\frac{7\times 6!}{6!}=7$

$\binom 72=\frac{7!}{2!(7-2)!}=\frac{7\times 6 \times 5!}{2!\times 5!}=21$

$\binom 73=\frac{7!}{3!(7-3)!}=\frac{7\times 6 \times 5 \times 4!}{3!\times 4!}=35$

$\binom 74=\frac{7!}{4!(7-4)!}=\frac{7\times 6\times 5 \times 4!}{4!\times 3!}=35$

$\binom 75=\frac{7!}{5!(7-5)!}=\frac{7\times 6 \times 5!}{2!\times 5!}=21$

$\binom 76=\frac{7!}{6!(7-6)!}=\frac{7\times 6!}{6!}=7$

$\binom 77=\frac{7!}{7!(7-7)!}=\frac{7!}{7!}=1$

Finally:

$(x+y)^7=x^7+7 x^6y+21 x^5y^2+35 x^4y^3+35 x^3y^4+21 x^2y^5+7 xy^6+y^7$

We will combine this information with the De Moivre theorem to find few trigonometric formulas.

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