Usuario discusión:Cesar

Sean $w,z\in {C}$ Demostrar que:

a)$||z||=||{\overline {z}}||$

Demostración

Sea ${\overline {z}}=a-ib\qquad y\qquad z=a+ib$ entonces

$||{\overline {z}}||={\sqrt {a^{2}+(-b)^{2}}}\qquad y\qquad ||z||={\sqrt {a^{2}+b^{2}}}$

$\therefore \qquad ||z||=||{\overline {z}}||$

b) $\qquad ||zw||=||z||||w||$

Demostración

$||zw||^{2}=(zw)({\overline {zw}})$

$||zw||^{2}=(z{\overline {w}})(w{\overline {z}})$

$\qquad ||zw||^{2}=||z||^{2}||w||^{2}$

$\therefore \qquad ||zw||=||z||||w||$

c) $\qquad ||z+w||\leq ||z||+||w||$

Demostración

$||z+w||^{2}=(z+w)({\overline {w+z}})$

$||z+w||^{2}=(z+w)({\overline {w}}+{\overline {z}})$

$||z+w||^{2}=z{\overline {z}}+z{\overline {w}})+w{\overline {z}}+w{\overline {w}}$

$\qquad ||z+w||^{2}=||z||^{2}+2Re(zw)+||w||^{2}$

$\qquad ||z+w||^{2}\leq ||z||^{2}+2||z||||w||+||w||^{2}$

$\qquad ||z+w||^{2}\leq (||z||+||w||)^{2}$

$\therefore \qquad ||z+w||\leq ||z||+||w||$

d) $||z-w||\geq ||z||-||w||$

Demostración

$\qquad ||z||=||(z-w)+w||$

$||z||\leq ||z-w||+||w||$

$\therefore \qquad ||z||-||w||\leq ||z-w||$

Si z=a+ib, demuestre que $Re(z)={\frac {z+{\overline {z}}}{2}}\qquad y\qquad Im(z)={\frac {z-{\overline {z}}}{2i}}$

Demostración

a) $\qquad {\frac {z+{\overline {z}}}{2}}={\frac {a+ib+(a-ib)}{2}}$

${\frac {z+{\overline {z}}}{2}}={\frac {2a}{2}}$

$\therefore \qquad {\frac {z+{\overline {z}}}{2}}=Re(a)$

b)$\qquad {\frac {z-{\overline {z}}}{2i}}={\frac {a+ib-(a-ib)}{2i}}$

${\frac {z+{\overline {z}}}{2i}}={\frac {2ib}{2i}}$

$\therefore \qquad {\frac {z+{\overline {z}}}{2i}}=Im(z)$

LEYES DE MORGAN

1) $(A\cup B)^{C}=A^{C}\cap B^{C}$

Demostración

Sea un x arbitrario del conjunto universal, entonces:

$x\in (A\cup B)^{C}\Leftrightarrow x\notin (A\cup B)$

$\Leftrightarrow \neg (x\in (A\cup B))$

$\Leftrightarrow \neg (x\in A\vee x\in B)$

$\Leftrightarrow x\notin A\wedge x\notin B$

$\Leftrightarrow (x\in A^{C})\wedge (x\in B^{C})$

$\Leftrightarrow x\in (A^{C}\cap B^{C})$

al ser x arbitrario

$\forall x:x\in ((A\cup B)^{C})\Leftrightarrow x\in (A^{C}\cap B^{C})$

$\therefore (A\cup B)^{C}=A^{C}\cap B^{C}$