Answer to Question #138035 in Real Analysis for Mubina

Question #138035

Prove that the set of complex numbers is field.


1
Expert's answer
2020-10-13T19:38:49-0400

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\u00d7y)=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 \u00d7 w = (x,y) \u00d7 (u,v) = (x\u00d7u - y\u00d7v, x\u00d7v + y\u00d7u)"

Now, for the first case; (when "(+)" and "(\u00d7)" are associative). It is simple for addition, but multiplication requires illustrations that actually "a\u00d7(b\u00d7c)=(a\u00d7b)\u00d7c"

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 "(\u00d7)", we can prove that

"a+b = b+a" and "a\u00d7b = b\u00d7a"

"(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, "(\u00d7)"


For the distributive law "a\u00d7(b+c)=(a\u00d7b)+(a\u00d7c)" we have;

"a\u00d7(b+c)=(x_1+y_1) \u00d7 ([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\u2212y_1y_2)+(x_1y_2+y_1x_2)]+[(x_1x_3\u2212y_1y_3)+(x_1y_3+y_1x_3)]"

"=(a\u00d7b)+(a\u00d7c)"


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\u00d71=(x_1+y_1)(1+0i)"

"=(x_1+x_1\u00d70i)+(y_1+y_1\u00d70)i"

"=x_1+y_1i"

"=z_1"

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

"z_1" which is "\u2212z1=\u2212x\u2212yi"

That is "z_1+(\u2212z_1)=(x_1+y_1i)+(\u2212x_1\u2212y_1i)"

"=(x_1\u2212x_1)+(y_1\u2212y_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\u2212y_1i}{x_1^2+y_1^2}"

Now "z_1\u00d7z_1^{\u22121} = (x_1+y_1i) \u00d7 (\\frac{x_1\u2212y_1i}{x_1^2+y_1^2})"


"=\\frac{(x_1+y_1i)(x_1\u2212y_1i)}{x_1^2+y_1^2}"


"= \\frac{x_1^2\u2212x_1y_1i + x_1y_1i\u2212y_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.


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