Question

Prove that the set of complex numbers is field.

Answer 1

SOLUTION

The set of complex numbers, $C$ with addition and multiplication is defined as a field comprising of additive and multiplicative credentials $(0,0)$ and $(1,0).$

With the complex number $i = (0,1)$, it is possible to interchangeably list each complex number as

$z=(x,y) = x+(i×y)=x+iy$

Now, in order to prove that set of complex numbers is field, we need to find the field axioms for the definitions of addition and multiplication i.e.

$z + w = (x,y) + (u,v) = (x+u, y+v)$

$z × w = (x,y) × (u,v) = (x×u - y×v, x×v + y×u)$

Now, for the first case; (when $(+)$ and $(×)$ are associative). It is simple for addition, but multiplication requires illustrations that actually $a×(b×c)=(a×b)×c$

Now, let $a=(x_1,y_1), b=(x_2,y_2), and, c=(x_3,y_3)$ such that;

$a.(b.c) = (x_1, y_1).((x_2, y_2). (x_3, y_3))$ $= (x_1, y_1). (x_2x_3 - y_2y_3, x_2x_3 + y_2y_3)$

$=(x_1(x_2x_3 - y_2y_3) - y_1(x_2x_3 + y_2y_3), x_1(x_2x_3 + y_2y_3) + y_1(x_2x_3 - y_2y_3))$

$=(x_1x_2x_3 - x_1y_2y_3 - y_1x_2x_3 - y_1y_2y_3, x_1x_2x_3 + x_1y_2y_3 + y_1x_2x_3 - y_1y_2y_3)$

$= (x_3(x_1x_2 - y_1y_2) - y_3(x_1y_2 + y_1x_2), x_3(x_1y_2 + y_1x_2)+ y_3(x_1x_2 - y_1y_2))$

$= (x_1x_2 - y_1y_2, x_1y_2 + y_1x_2). (x_3,y_3)$

$=((x_1,y_1).(x_2,y_2)) .(x_3,y_3)$

$=(a.b).c$

For commutative of both $(+)$ and $(×)$, we can prove that

$a+b = b+a$ and $a×b = b×a$

$(x_1+y_1)+(x_2+y_2) =(x_1 +x_2) + (y_1 +y_2)$

$=(x_2+x_1) + (y_2+y_1)$

$=(x_2+ y_2) +(x_1+y_1)$

$=b+a$

This also applies to the multiplication, $(×)$

For the distributive law $a×(b+c)=(a×b)+(a×c)$ we have;

$a×(b+c)=(x_1+y_1) × ([x_2+x_3]+[y_2+y_3])$

$=x_1[x_2+x_3]+x_1[y_2+y_3]+y_1[x_2+x_3]+y_1[y_2+y_3]$

$=[(x_1x_2−y_1y_2)+(x_1y_2+y_1x_2)]+[(x_1x_3−y_1y_3)+(x_1y_3+y_1x_3)]$

$=(a×b)+(a×c)$

For the additive identity, $(0,0)$ we have;

$0=0+0i.$

That is $z_1+0=(x_1+y_1i)+(0+0i)$

$=(x_1+0)+(y_1+0)i$

$=x_1+y_1i$

$=z_1$

For the multiplicative identity $(1,0)$ we have

$1=1+0i$

That is $z_1×1=(x_1+y_1)(1+0i)$

$=(x_1+x_1×0i)+(y_1+y_1×0)i$

$=x_1+y_1i$

$=z_1$

For the additive inverse of $(x,y)$ gives $(-x,-y)$ i.e. Considering;

$z_1$ which is $−z1=−x−yi$

That is $z_1+(−z_1)=(x_1+y_1i)+(−x_1−y_1i)$

$=(x_1−x_1)+(y_1−y_1)i$

$=0+0i$

$=0$

For the multiplicative inverse to $(x,y)$ is $(\frac{x}{(x2+y2)}, -\frac{y}{(x2+y2) })$ i.e.

considering complex number $z_1$ where $z_1 {=}\mathllap{/\,}0$

$z_1^{-1} = \frac{x_1−y_1i}{x_1^2+y_1^2}$

Now $z_1×z_1^{−1} = (x_1+y_1i) × (\frac{x_1−y_1i}{x_1^2+y_1^2})$

$=\frac{(x_1+y_1i)(x_1−y_1i)}{x_1^2+y_1^2}$

$= \frac{x_1^2−x_1y_1i + x_1y_1i−y_1i^2}{x_1^2+y_1^2}$

$= \frac{x_1^2+y_1^2}{x_1^2+y_1^2}$

$=1$

Therefore, Since the set of complex numbers $C$ used in the proof above

 satisfies the axioms working with the addition and multiplication, it can therefore be concluded that $C$ is a field.