TITLE: Obtain magnitude of square-rooted complex number
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: The square root of a complex number is not well defined; however, if we consider a complex number $z$ such that $z^2=w$, then
$$
|w|=|z^2|=|z|^2
$$
so
$$
|z|=\sqrt{|w|}
$$
(and this is well defined, since we are talking of nonnegative real numbers).
Your situation is exactly this one, with a slight complication. However, since
$$
|w^{-1}|=|w|^{-1}
$$
for any $w\ne0$, you know that, when $w=\alpha+i\beta$ and $z^2=w^{-1}$
that
$$
|z|=\sqrt{|w^{-1}|}=\sqrt{|w|^{-1}}=\frac{1}{\sqrt{|w|}}
$$
because of the arguments above. So
$$
|z|=\frac{1}{\sqrt{|w|}}=\frac{1}{\sqrt{\sqrt{\alpha^2+\beta^2}}}
=\frac{1}{\sqrt[4]{\alpha^2+\beta^2}}
$$