How to compute derive a complex number from a square root of another complex number?

Question

In plane wave in a lossy medium, the complex propagative constant is given to be

$$\gamma = \alpha\ + j\beta $$ $$= j \omega \sqrt {\mu \epsilon_c}$$ $$= j \omega \sqrt { \mu \left(\epsilon - j\frac{\sigma}{\omega}\right) }$$

Which will eventually result in a square root of a complex number like

$$\gamma = j\omega \sqrt {X+jY}$$

how do I proceed on to find $\gamma$ in terms of $\alpha$ and $\beta$ ?

Answer 1

Note that if $(X+jY)^2=a+jb$ with $a:=-\tfrac{\alpha}{\omega^2},\,b:=-\tfrac{\beta}{\omega^2}$ then $X^2-Y^2=a,\,2XY=b$. Squaring, suming and square-rooting, $X^2+Y^2=\sqrt{a^2+b^2}$. Averaging $X^2\pm Y^2$, $X^2=\frac{\sqrt{a^2+b^2}+a}{2}$ and $Y^2=\frac{\sqrt{a^2+b^2}-a}{2}$. While $X^2$ ($Y^2$) is consistent with two values of $X$ ($Y$), only two solutions exist, not four; the other two have a sign error in the value of $XY$.

Answer 2

Let $\sqrt{a+ib}=x+iy$, or $a+ib=(x+iy)^2=x^2-y^2+i2xy$.

So after identification,

$$ax^2=x^4-x^2y^2=x^4-\frac{b^2}4$$

and solving that biquadratic equation,

$$x=\pm\sqrt{\frac{\sqrt{a^2+b^2}+a}2}$$ and $$y=\frac b{2x}.$$

Answer 3

Let $z$ be the complex radicand and render

$(z+|z|)^2=z^2+2z|z|+|z|^2=z^2+2z|z|+z\overline z=z(2|z|+2\Re z)$

Thus for $z$ not a negative real number so that $|z|+\Re z>0$:

$z+|z|=\pm(\sqrt z)(\sqrt{2(|z|+\Re z)})$

$\color{blue}{\sqrt z=\pm\dfrac{z+|z|}{\sqrt{2(|z|+\Re z)}}}$