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I'm stuck at doing this problem. It's actually a problem related to circuit analysis. I'm given this equation: $$\frac{(R_3+\frac{1}{\omega C}I)*(\omega Li)}{(R_3+\frac{1}{\omega Ci})+(\omega Li)}$$ Now to rationalize it I will multiply by $$\frac{R_3+\frac{1}{\omega C}i-\omega Li}{R_3+\frac{1}{\omega C}i-\omega Li}$$ $$\frac{R_3+\frac{1}{\omega C}i*\omega Li}{R_3+\frac{1}{\omega C}i+\omega Li}*\frac{R_3+\frac{1}{\omega C}i*\omega Li}{R_3+\frac{1}{\omega C}i+\omega Li}$$ After I do the actions I get: $$\frac{(R_3+\frac{1}{\omega C}i)*(\omega Li)*R_3+\frac{1}{\omega C}i-\omega Li}{(R_3+\frac{1}{\omega C}i)^2+(\omega Li)^2}$$ The problem is that I've to get rid of the imaginary part from the denominator.Any ideas how I can accomplish this?

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Well, as an electrical engineer you'll write (where $\text{j}^2=-1$):

$$\underline{\text{Z}}_{\space\text{in}}=\frac{\left(\underline{\text{Z}}_{\space\text{R}}+\underline{\text{Z}}_{\space\text{C}}\right)\cdot\underline{\text{Z}}_{\space\text{L}}}{\underline{\text{Z}}_{\space\text{R}}+\underline{\text{Z}}_{\space\text{C}}+\underline{\text{Z}}_{\space\text{L}}}=\frac{\left(\text{R}+\frac{1}{\text{j}\omega\text{C}}\right)\cdot\text{j}\omega\text{L}}{\text{R}+\frac{1}{\text{j}\omega\text{C}}+\text{j}\omega\text{L}}=\frac{\left(\text{R}-\frac{\text{j}}{\omega\text{C}}\right)\cdot\text{j}\omega\text{L}}{\text{R}-\frac{\text{j}}{\omega\text{C}}+\text{j}\omega\text{L}}=$$ $$\frac{\text{R}\cdot\text{j}\omega\text{L}-\frac{\text{j}}{\omega\text{C}}\cdot\text{j}\omega\text{L}}{\text{R}-\frac{\text{j}}{\omega\text{C}}+\text{j}\omega\text{L}}=\frac{\frac{\omega\text{L}}{\omega\text{C}}+\text{R}\omega\text{L}\cdot\text{j}}{\text{R}+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\cdot\text{j}}=\frac{\frac{\text{L}}{\text{C}}+\text{R}\omega\text{L}\cdot\text{j}}{\text{R}+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\cdot\text{j}}=$$ $$\frac{\left(\frac{\text{L}}{\text{C}}+\text{R}\omega\text{L}\cdot\text{j}\right)\cdot\overline{\text{R}+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\cdot\text{j}}}{\left(\text{R}+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\cdot\text{j}\right)\cdot\overline{\text{R}+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\cdot\text{j}}}=$$ $$\frac{\left(\frac{\text{L}}{\text{C}}+\text{R}\omega\text{L}\cdot\text{j}\right)\cdot\left(\text{R}-\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\cdot\text{j}\right)}{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)^2}=$$ $$\frac{\frac{\text{L}}{\text{C}}\cdot\text{R}-\frac{\text{L}}{\text{C}}\cdot\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\cdot\text{j}+\text{R}\omega\text{L}\cdot\text{j}\cdot\text{R}-\text{R}\omega\text{L}\cdot\text{j}\cdot\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\cdot\text{j}}{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)^2}=$$ $$\frac{\frac{\text{L}}{\text{C}}\cdot\text{R}+\text{R}\omega\text{L}\cdot\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)+\left(\text{R}^2\omega\text{L}-\frac{\text{L}}{\text{C}}\cdot\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\right)\cdot\text{j}}{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)^2}\tag1$$

Because you're analysing a parallel circuit of a series resistor and capacitor and parallel to the series resistor and capacitor a coil.

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use that $$\frac{R_3-\left(\frac{1}{\omega C}\omega L\right)}{R_3+i\left(\frac{1}{\omega C}-\omega L\right)}$$ and multiply numerator and denominator by $$R_3-i\left(\frac{1}{\omega C}-\omega L\right)$$

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  • $\begingroup$ the imaginary part is in front of the whole fraction. $\endgroup$ – Deyan Georgiev Jan 10 '18 at 19:13

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