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How do the 2 complex values equate to 3?
\[\large\rm \frac{3}{1+3i \omega}+\frac{9i \omega}{1+3i \omega}\quad=\quad\frac{3+9i \omega}{1+3i \omega}\]From here, factor a 3 out of each term in the numerator,\[\large\rm =\frac{3(1+3i \omega)}{1+3i \omega}\]
\[\large\rm =\frac{3\cancel{(1+3i \omega)}}{\cancel{1+3i \omega}}\]Understand? :)
are you finding the transfer function at resistor ?
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