Mostre que

a ) |overset(-)(z)| = | z |

b ) |z w| = |z| | w |

c ) |(z)/(w)| = (| z |)/(| w |)

d ) |z + w|^2 = |z|^2 + |w|^2 + 2 R e ( z overset(-)(w) )

|z - w|^2 = |z|^2 + |w|^2 - 2 R e ( z overset(-)(w ))

Passo 1

Para mostrar tais itens vamos tomar z = a + b i e w = c + d i

|z| = sqrt(a^2 + b^2)

|w| = sqrt(c^2 + d^2)

Letra a)

|overset(-)(z)| = |a - b i| = sqrt(a^2 + b^2) = | z |

Letra b)

|z w| = |a c - b d + i (a d + b c)| = sqrt((a c - b d)^2 + (a d + b c)^2 ) =

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

= sqrt(a^2 c^2 + b^2 d^2 + a^2 d^2 + b^2 c^2) =

= sqrt(( a^2 + b^2 ) ( c^2 + d^2 )) = |z| | w |

Passo 2

Letra c)

|(z)/(w)| = |(z overset(-)(w))/(w overset(-)(w))|

Pela propriedade da b:

|(z overset(-)(w))/(w overset(-)(w))| = |(1)/(w overset(-)(w))| |z| |overset(-)(w)| = |(1)/(w)| |(1)/(overset(-)(w))| |overset(-)(w)| |z| = (| z |)/(| w |)

Passo 3

Letra d)

|z + w| = |a + c + i (b + d)| = sqrt((a + c)^2 + (b + d)^2)

|z + w|^2 = (a + c)^2 + (b + d)^2

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

Obviamente que a^2 + b^2 + c^2 + d^2 = |z|^2 + |w|^2

Basta analisarmos o que é o resto

z overset(-)(w) = (a + b i) (c + d i) = a c + a d i + b i c - b d = a c - b d + i ( a d + b i )

Logo R e (z overset(-)(w)) = a c - b d

|z + w|^2 = |z|^2 + |w|^2 + 2 R e (z overset(-)(w))

A outra é análoga:

|z - w| = |a - c + i (b - d)| = sqrt((a + c)^2 + (b - d)^2)

|z - w|^2 = (a - c)^2 + (b - d)^2

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

|z|^2 + |w|^2 - 2 R e ( z overset(-)(w ))

Resposta

Demonstração