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Sea  \frac{x-iy}{x+iy}=a+ib. Probar que a^{2}+b^{2}=1

1 Respuesta

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\frac{x-iy}{x+iy}=\frac{x-iy}{x+iy}\frac{x-iy}{x-iy}=\frac{(x-iy)^{2}}{(x+iy)(x-iy)}=\frac{(x^{2}-y^{2})-2xyi}{x^{2}+y^{2}}

= \frac{x^{2}-y^{2}}{x^{2}+y^{2}}-i\left ( \frac{2xy}{x^{2}+y^{2}} \right )

Asì

a= \frac{x^{2}-y^{2}}{x^{2}+y^{2}}

b=-\left ( \frac{2xy}{x^{2}+y^{2}} \right )

\Rightarrow

a^{2}=\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )^{2}    y    b^{2}= \left (- \frac{2xy}{x^{2}+y^{2}} \right )^{2}

a^{2}+b^{2}=\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )^{2}+ \left (- \frac{2xy}{x^{2}+y^{2}} \right )^{2}

=\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )^{2}+\frac{(2xy)^{2}}{(x^{2}+y^{2})^{2}}

=\frac{1}{x^{2}+y^{2}}\left ( (x^{2}-y^{2})^{2}+(2xy)^{2} \right )

=\frac{(x^{2}+y^{2})^{2}}{(x^{2}+y^{2})^{2}}

=1

Por lo tanto a^{2}+b^{2}=1

respondido por Rocio Oct 6
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