Compleja:ej-cap1.1

1.1.1

1. Demuestre que el producto de números complejos cumple con la ley asociativa

Sean $z=a+ib,\quad w=c+id,\quad s=e+if\,\quad$ con $\quad a,b,c,d,e,f\in {\mbox{R}}$

Por demostrar $(zw)s=z(ws)\,$

$(zw)s=[(a+ib)(c+id)](e+if)=[(ac-bd)+i(bc+ad)](e+if)\,$

$=[e(ac-bd)-f(bc+ad)]+i[e(bc+ad)+f(ac-bd)=(ace-bde-bcf-adf)+i(bce+ade+acf-bdf)\,$

Por otra parte

$z(ws)=(a+ib)[(c+id)(e+if)]=(a+ib)[(ce-df)+i(de+cf)]\,$

$=[a(ce-df)-b(de+cf)]+i[b(ce-df)+a(de+cf)]=(ace-bde-bcf-adf)+i(bce+ade+acf-bdf)\,$

Entonces se cumple $(zw)s=z(ws)\,$.

1.1.2

1. Demuestre que $\left|{\frac {z}{w}}\right|={\frac {\left|z\right|}{\left|w\right|}}$

Sean $z=a+ib\quad y\quad w=c+id\,$

$\left|{\frac {z}{w}}\right|=\left|{\frac {a+ib}{c+id}}\right|=\left|{\frac {(a+ib)(c-id)}{(c+id)(c-id)}}\right|$

$=\left|{\frac {(ac+bd)+i(bc-ad)}{c^{2}+d^{2}}}\right|={\sqrt {{\bigg (}{\frac {ac+bd}{c^{2}+d^{2}}}{\bigg )}^{2}+{\bigg (}{\frac {bc-ad}{c^{2}+d^{2}}}{\bigg )}^{2}}}={\frac {1}{c^{2}+d^{2}}}{\sqrt {(ac+bd)^{2}+(bc-ad)^{2}}}$

$={\frac {1}{c^{2}+d^{2}}}{\sqrt {a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}}}={\sqrt {{\Bigg [}{\frac {a^{2}(c^{2}+d^{2})}{(c^{2}+d^{2})^{2}}}{\Bigg ]}+{\Bigg [}{\frac {b^{2}(c^{2}+d^{2})}{(c^{2}+d^{2})^{2}}}{\Bigg ]}}}={\sqrt {\frac {a^{2}+b^{2}}{c^{2}+d^{2}}}}$

Por otra parte

${\frac {\left|z\right|}{\left|w\right|}}={\frac {\left|a+ib\right|}{\left|c+id\right|}}={\sqrt {\frac {a^{2}+b^{2}}{c^{2}+d^{2}}}}=\left|{\frac {z}{w}}\right|$

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2. Exprese ${\overline {\left({\frac {\left(2+3i\right)^{2}}{4+i}}\right)}}$de la forma $x+iy$

aqui ya comienza su demostración ...

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etc ...

1.1.4

--mfg-wiki 21:27 25 sep 2009 (UTC)