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?
$\endgroup$ 28 Answers
$\begingroup$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 $|z| = \sqrt{z\bar{z}}$.
$\endgroup$ $\begingroup$Assume $\alpha,\beta\in\mathbb{R}$:
$$\left|\frac{1}{\sqrt{\alpha+i\beta}}\right|=\frac{\left|1\right|}{\left|\sqrt{\alpha+i\beta}\right|}=\frac{1}{\left|\left(\alpha+i\beta\right)^{\frac{1}{2}}\right|}=\frac{1}{\left|\alpha+i\beta\right|^{\frac{1}{2}}}=$$ $$\frac{1}{\left(\sqrt{\alpha^2+\beta^2}\right)^{\frac{1}{2}}}=\frac{1}{\left(\left(\alpha^2+\beta^2\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}}=\frac{1}{\left(\alpha^2+\beta^2\right)^{\frac{1}{4}}}=\frac{1}{\sqrt[4]{\alpha^2+\beta^2}}$$
$\endgroup$ $\begingroup$$$|z|=|(\alpha+i\beta)^{-\frac{1}{2}}|=|\alpha+i\beta|^{-\frac{1}{2}}=\left(\sqrt{\alpha^2+\beta^2}\right)^{-\frac{1}{2}}=\frac{1}{\sqrt[4]{\alpha^2 + \beta^2}}$$
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.
$\endgroup$ $\begingroup$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 $$ |z|=\frac{1}{\sqrt{|w|}}=\frac{1}{\sqrt{\sqrt{\alpha^2+\beta^2}}} =\frac{1}{\sqrt[4]{\alpha^2+\beta^2}} $$
$\endgroup$ $\begingroup$You could also say: $$\alpha +\beta i=re^{i\theta },$$ for some $r$ and $\theta$. This means that $$\frac{1}{\sqrt {\alpha +\beta i}}=\left (re^{i\theta} \right)^{-\frac 12}$$ has magnitude $r^{-\frac12}$, where $r$ is the magnitude of $\alpha +\beta i$. This means $$r^{-\frac12} =\left(\sqrt{\alpha^2+\beta^2}\,\,\right)^{-\frac12}=\frac 1 {\sqrt[4]{\alpha^2+\beta^2}}.$$
$\endgroup$ $\begingroup$$$ |z| = \sqrt{zz^*} $$ so we have $$ \sqrt{\frac{1}{\sqrt{\alpha+i\beta}}\frac{1}{\sqrt{\alpha-i\beta}}}=\sqrt{\frac{1}{\sqrt{(\alpha+i\beta)(\alpha-i\beta)}}} = ? $$
$\endgroup$ 1 $\begingroup$$w\in \mathbb C$ lies on a circle of radius $|w|$ centered at $0$, and every value of $w^p$ (for $p\in \mathbb R$) lies on a circle of radius $|w|^p$ centered at $0$.
So both values of $(\alpha+i \beta)^{-\frac12}$ lie on a circle of radius $|\alpha + i\beta|^{-\frac12} = (\alpha^2+\beta^2)^{-\frac14}$ centered at $0$.
$\endgroup$ $\begingroup$The magnitude of a complex number is defined like this: $|z|^2 = z\bar{z}$.
In your particular case, that means: $\frac{1}{\sqrt{\alpha + i\beta}}\frac{1}{\sqrt{\alpha - i\beta}} = \frac{1}{\sqrt{\alpha^2 + \beta^2}} = |z|^2$
What did I do here? Just notice that $\sqrt{a}\sqrt{b} = \sqrt{ab}$
and, since $i^2 = -1$ you can work this out as: $(a+ib)(a-ib) = a^2 - aib + aib - i^2b^2 = a^2 + b^2$
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