Obtain magnitude of square-rooted complex number

If you convert the number to the complex exponential form, the solution is easy.

Let $s = \alpha + \beta i = r e^{\theta i}$, then $z = s^{-\frac{1}{2}} = r^{-\frac{1}{2}} e^{-\frac{\theta}{2}i}$. The conjugate (written with an overbar) of a complex exponential $re^{\theta i}$ is just $re^{-\theta i}$, so calculating $z\bar{z}$ leads to the exponential terms cancelling and leaves $z\bar{z} = r^{-1}$. Now $r = \sqrt{\alpha^2 + \beta^2}$ and you need $|z| = \sqrt{z\bar{z}}$.

Assume $\alpha,\beta\in\mathbb{R}$:

$$\left|\frac{1}{\sqrt{\alpha+i\beta}}\right|=\frac{\left|1\right|}{\left|\sqrt{\alpha+i\beta}\right|}=\frac{1}{\left|\left(\alpha+i\beta\right)^{\frac{1}{2}}\right|}=\frac{1}{\left|\alpha+i\beta\right|^{\frac{1}{2}}}=$$ $$\frac{1}{\left(\sqrt{\alpha^2+\beta^2}\right)^{\frac{1}{2}}}=\frac{1}{\left(\left(\alpha^2+\beta^2\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}}=\frac{1}{\left(\alpha^2+\beta^2\right)^{\frac{1}{4}}}=\frac{1}{\sqrt[4]{\alpha^2+\beta^2}}$$

$$|z|=|(\alpha+i\beta)^{-\frac{1}{2}}|=|\alpha+i\beta|^{-\frac{1}{2}}=\left(\sqrt{\alpha^2+\beta^2}\right)^{-\frac{1}{2}}=\frac{1}{\sqrt[4]{\alpha^2 + \beta^2}}$$

By the way, you'd need te define what you mean by $\sqrt{\alpha+i\beta}$ or $(\alpha+i\beta)^{-\frac{1}{2}}$ because there are $2$ complex numbers satisfying $z^2=\alpha+i\beta$. They have the same modulus but still. Once you gave some meaning to that, you can show the property on modulus that I used.