Let's consider the following quantity:
$$K = \sqrt{\frac{c}{a + i b}}$$
where $i$ is the imaginary unit and $a, b, c \in \mathbb{R}$.
Is it possible to express it in the complex exponential form? That is
$$K = \alpha e^{i\beta}$$
It could be not immediate to compute $\alpha$ as $\sqrt{(\mathrm{Re} \{ K \})^2 + (\mathrm{Im} \{ K \})^2}$ and $\beta$ as $\arctan$ of $\mathrm{Im} \{ K \} / \mathrm{Re} \{ K \}$. Does maybe exist a more straightforward way in such a case?
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Let $K=\sqrt \frac{c}{a+ib}=x+iy$
On squaring both the sides -
$\frac{c}{a+ib}=(x+iy)^2$
$\frac{c}{a+ib}=x^2-y^2+2ixy$
Now multiply and divide LHS by $(a-ib)$
$\frac{c(a-ib)}{a^2-b^2}=x^2-y^2+2ixy$
On comparing real and imaginary coefficients, you can get values of $x$ and $y$ in terms of $a,b$ and $c$.
Then let $\sqrt {x^2+y^2}=\alpha$ and $\cos\beta = \frac{x}{\alpha} $.
Now,
$K=x+iy=\alpha(\frac{x}{\alpha}+i\frac{y}{\alpha})$
$K=\alpha e^{i\beta}$
$\alpha = \sqrt {x^2+y^2}$
$\beta = \arccos \frac {x}{\sqrt {x^2+y^2}}$
I didn't quite solve it, but here are some thoughts to get you inspired: $$x + iy = re^{i \theta} = K = \sqrt{ \frac{c}{a + bi}} = \sqrt{ \frac{c(a - bi)}{a^2 + b^2}} = \sqrt{ \frac{ca}{a^2 + b^2} - i\frac{cb}{a^2 + b^2}}$$ We know that $tan(\theta) = \frac{x}{y}$ and $r = |z| = \sqrt{ x^2 + y^2}$. Then$$ (x + iy)^2 = \bigg(\sqrt{ \frac{ca}{a^2 + b^2} - i\frac{cb}{a^2 + b^2}}\bigg)^2$$ or $$ x^2 - y^2 + 2ixy = \frac{ca}{a^2 + b^2} - i\frac{cb}{a^2 + b^2}$$ and last you need to solve for x and y given $$x^2 - y^2 = \frac{ca}{a^2 + b^2}$$ and $$2yx = - \frac{cb}{a^2 + b^2}$$ which I have to admit must be quite cumbersome. However, once you compute x and y, just plug them in $tan(\theta) = \frac{x}{y}$ and $r = |z| = \sqrt{ x^2 + y^2}$ to find $\theta$ and $r$, and you are done.
Alternatively, you can start by squaring both sides: $$ r^2e^{i 2\theta} = \bigg(\sqrt{ \frac{ca}{a^2 + b^2} - i\frac{cb}{a^2 + b^2}}\bigg)^2$$ or $$ r^2e^{i 2\theta} = \frac{ca}{a^2 + b^2} - i\frac{cb}{a^2 + b^2} $$ with $ r^2 = u^2 + v^2$, where u,v are the real and imaginary parts respectively in the above expression, and play around with the argument geometrically until find $\theta$, but I think that will be harder.
Good luck!