Functie inversabila
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BogdanCNFB
- Thales
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Functie inversabila
Fie \( \omega\in C-R,|\omega|\neq 1 \). Sa se demonstreze ca functia \( f:C\rightarrow C,f(z)=z+\omega\overline{z} \) este inversabila si sa se calculeze \( f^{-1} \).
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Laurian Filip
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\( w=ki \) , \( k \in \mathbb{R} \)
fie \( z=a+b \)i
\( f(z)=a+bi+ki(a-bi)=a+bk+(b+ak)i
\)
fie \( x=u+vi \) un numar complex
\( \begin{cases} u=a+bk \\ v=b+ak \end{cases} \)
pt \( |k| \neq 1 \) acest sistem are solutie unica (\( \frac{1}{k} \neq k \)), deci functia e bijectiva , deci e si inversabila.
\( u-vk=a-ak^2 \)
\( a=\frac{u-vk}{1-k^2} \)
\( b=\frac{v-uk}{1-k^2} \)
\( f^{-1}(x)=\frac{u-vk}{1-k^2}+\frac{v-uk}{1-k^2}i \)
fie \( z=a+b \)i
\( f(z)=a+bi+ki(a-bi)=a+bk+(b+ak)i
\)
fie \( x=u+vi \) un numar complex
\( \begin{cases} u=a+bk \\ v=b+ak \end{cases} \)
pt \( |k| \neq 1 \) acest sistem are solutie unica (\( \frac{1}{k} \neq k \)), deci functia e bijectiva , deci e si inversabila.
\( u-vk=a-ak^2 \)
\( a=\frac{u-vk}{1-k^2} \)
\( b=\frac{v-uk}{1-k^2} \)
\( f^{-1}(x)=\frac{u-vk}{1-k^2}+\frac{v-uk}{1-k^2}i \)