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I was solving exercises related to operational amplifiers and I got to a part where I am not sure how I am getting the arg part of the result. Can someone please explain this to me. Thank you very much.

$$|H(j\omega)| = \left|-\frac{j\omega LR_2}{R_1(R_2+j\omega L)}\right| = \frac{\omega L\frac{R_2}{R_1}}{\sqrt{R^2_2+(\omega L)^2}}$$

$$\arg\{H(j\omega)\} = -\frac{\pi}{2} - \arctan\left(\frac{\omega L}{R_2}\right)$$

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    \$\begingroup\$ I have replaced your image with MathJax (edit may be pending). Please check that I have done so correctly and consider using MathJax in the future. \$\endgroup\$ Feb 20, 2022 at 21:14

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Well, we have:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=-\frac{\text{j}\omega\text{LR}_2}{\text{R}_1\left(\text{R}_2+\text{j}\omega\text{L}\right)}\tag1$$

Using the principal value of the argument we know that we can write:

\begin{equation} \begin{split} \arg\left(\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)&=\arg\left(-\frac{\text{j}\omega\text{LR}_2}{\text{R}_1\left(\text{R}_2+\text{j}\omega\text{L}\right)}\right)\\ \\ &=\arg\left(-1\right)+\arg\left(\frac{\text{j}\omega\text{LR}_2}{\text{R}_1\left(\text{R}_2+\text{j}\omega\text{L}\right)}\right)\\ \\ &=\arg\left(-1\right)+\arg\left(\text{j}\omega\text{LR}_2\right)-\arg\left(\text{R}_1\left(\text{R}_2+\text{j}\omega\text{L}\right)\right)\\ \\ &=\underbrace{\arg\left(-1\right)}_{=\space\pi}+\underbrace{\arg\left(\text{j}\omega\text{LR}_2\right)}_{=\space\frac{\pi}{2}}-\left(\underbrace{\arg\left(\text{R}_1\right)}_{=\space0}+\arg\left(\text{R}_2+\text{j}\omega\text{L}\right)\right)\\ \\ &=\pi+\frac{\pi}{2}-\left(0+\arctan\left(\frac{\omega\text{L}}{\text{R}_2}\right)\right)\\ \\ &=\pi+\frac{\pi}{2}-\arctan\left(\frac{\omega\text{L}}{\text{R}_2}\right)\\ \\ &=\frac{3\pi}{2}-\arctan\left(\frac{\omega\text{L}}{\text{R}_2}\right) \end{split}\tag2 \end{equation}

Which is the same as your answer, because what happens when you add \$2\pi\$ on the right-hand side of your argument equation?

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