calculus - Obtain magnitude of square-rooted complex number

I would like to obtain the magnitude of a complex number of this form:

$$z = \frac{1}{\sqrt{\alpha + i \beta}}$$

By a simple test on WolframAlpha it should be

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

The fact is that if I try to cancel the root in the denominator I still have a troublesome expression at the numerator:

$$z = \frac{\sqrt{\alpha + i \beta}}{\alpha + i \beta}$$

And this alternative way seems unuseful too:

$$z = \left( \alpha + i \beta \right)^{-\frac{1}{2}}$$

If WolframAlpha gave the correct result, how to prove it?

Answer

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 $|z| = \sqrt{z\bar{z}}$.