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Suppose I have this equation:

$$H\left( e^{j\omega }\right) =\dfrac{1+e^{-j\omega }}{1-0.1e^{-j\omega }}$$

How do I find the magnitude exactly? I tried expanding it, but it seems to not work.

$$\frac{1+\cos\omega-j\sin\omega}{1-0.1\cos\omega+0.1j\sin\omega}$$

Finding the magnitude,

$$\sqrt{\frac{(1+\cos\omega-j\sin\omega)^2}{(1-0.1\cos\omega+0.1j\sin\omega)^2}}=\sqrt{\frac{(1+\cos\omega-j\sin\omega+\cos\omega+(\cos\omega)^2-j\sin\omega\cos\omega-(\sin\omega)^2}{(1-0.1\cos\omega+0.1j\sin\omega)^2}}$$

Now I do know that $$j\sin\omega\cos\omega$$ equates to 0 no matter what, leaving me with:

$$=\sqrt{\frac{(1+2\cos\omega-j\sin\omega+(\cos\omega)^2-(\sin\omega)^2}{(1-0.1\cos\omega+0.1j\sin\omega)^2}}$$

What am I supposed to simplify here?

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    $\begingroup$ Do you know how to compute the magnitude of a complex number $u=a+jb$? If not, you really need to learn it. Also, why do you think that $j\sin\omega\cos\omega$ is zero? It generally isn't. $\endgroup$
    – Matt L.
    Sep 5 '21 at 13:13
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The magnitude $|z|$ of a complex number $z=x+jy, \,\ x,y \in \mathbb{R}$ is \begin{equation} |z| = \sqrt{x^2 + y^2} \end{equation}

not \begin{equation} |z| = \sqrt{(x+jy)^{2}} \end{equation}

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you can rewrite the equation in this way \begin{align*} H(e^{j\omega}) &= \frac{1+e^{-j\omega}}{1-0.1e^{-j\omega}}\\ &= \frac {e^{-j\frac{\omega}{2}}\cdot(e^{j\frac{\omega}{2}}+ e^{-j\frac{\omega}{2}})} {0.9+0.1-0.1e^{-j\omega} }\\ &= \frac{e^{-j\frac{\omega}{2}}\cdot(e^{j\frac{\omega}{2}}+ e^{-j\frac{\omega}{2}})} {0.9+ 0.1 \cdot e^{-j\frac{\omega}{2}} \cdot (e^{j\frac{\omega}{2}}-e^{-j\frac{\omega}{2}})} \\ &= \frac{e^{-j\frac{\omega}{2}} 2\cdot \cos(\frac{\omega}{2})} {0.9+ e^{-j\frac{\omega}{2}} \cdot0.2j\cdot \sin(\frac{\omega}{2})} \end{align*} and the magnitude squared $|H(e^{j\omega})|^2$ is then \begin{equation*} |H(e^{j\omega})|^2 = \frac{2\cdot \cos(\frac{\omega}{2})} {(0.2\cdot \sin(\frac{\omega}{2})\sin(\omega)+ 0.9)^2 +0.04\cdot \sin(\frac{\omega}{2})^2\cos(\omega)^2} \end{equation*} note also that \begin{equation*} \sin(\omega)\cos(\omega)=\frac{\sin(2\omega)}{2} \end{equation*} which is equal to $0$ only for $\omega=\frac{k}{2}\pi$ for $k \in \mathbb{R}$

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