How to simplify these complex expressions? I have expressions:
$F(s)=\sqrt{\left[\left(a+b\right)s+c\right]\mp\sqrt{\left[\left(a-b\right)s+c\right]^{2}+d}}$
where $s$ is a complex number; and $a,b,c,d$ are reals such that
$a>0,\,b>0,\, a\neq b,\, c>0,\, d\geq0$. I am trying to figure out
whether
it is possible to simplify these expressions when $d=0$,
but I am afraid I have forgotten how to deal with the
constructs of the type $\sqrt{z^{2}}$ for the complex $z$ arising
here.
Can anybody help me please?
Leszek
Clarification 1:
I may be terribly uneducated, but what else can $\sqrt{z}$ mean, if not
"the square root of $z$"?
I never assume that the expressions of which the square roots are taken
are integers. Please read my description above.
I think my problem can be expressed in a simpler way: what is the result
of calculating $\sqrt{z^{2}}$ when $z$ is complex, and how this result depends on the actual form of $z$, as is described?
When $z$ is real, then $\sqrt{z^{2}}$ is equal to the absolute value of $z$.
Obviously I expect this is not the case when $z$ is complex.
Assuming a special case of real $s$  and $a > b$, we would obtain simplified expressions:
$F(s)=\sqrt{2bs}$ when there is "-" in the formula for $F(s)$,
and
$F(s)=\sqrt{2as+2c}$ when there is "+" in the formula for $F(s)$
I need to derive the counterparts of such simplified expressions obtainable for complex  $s$, both when $a > b$ and when $a < b$.
Leszek
 A: 
Clarification 1:
I may be terribly uneducated, but what else can $\sqrt{z}$ mean, if not "the square root of z"?
I never assume that the expressions of which the square roots are taken are integers. Please read my description above.
I think my problem can be expressed in a simpler way: what is the result of calculating $\sqrt{z^{2}}$ when z is complex, and how this result depends on the actual form of z, as is described?
When z is real, then $\sqrt{z^{2}}$ is equal to the absolute value of z. Obviously I expect this is not the case when z is complex.

$ \sqrt{z^{2}} $ is the unit square root of z.
However, if we say here that $ \sqrt{z^{2}} = \left( z^{2} \right)^{\frac{1}{2}} $, then it can also be $ \sqrt{z^{2}} = \pm z $ or just $ \sqrt{z^{2}} = z $.
$ \sqrt{z} $ as unit square root of z with $ z = \Re(z) + \Im(x) \cdot \mathrm{i} $ can be rewritten as $ \sqrt{z} = \sqrt{\frac{\Re(z) + |z|}{2}} + \operatorname{sign}(\Im(z)) \cdot \sqrt{\frac{-\Re(z) + |z|}{2}} \cdot \mathrm{i} $.
This in turn can be rewritten as:
$$
\begin{align*}
\sqrt{z} &= \sqrt{\frac{\Re(z) + |z|}{2}} + \frac{\Im(z)}{\Im(z)} \cdot \sqrt{\frac{-\Re(z) + |z|}{2}} \cdot \mathrm{i}\\
\sqrt{z} &= \sqrt{\frac{\Re(z) + \sqrt{\Re(z)^{2} + \Im(z)^{2}}}{2}} + \frac{\Im(z)}{|\Im(z)|} \cdot \sqrt{\frac{-\Re(z) + \sqrt{\Re(z)^{2} + \Im(z)^{2}}}{2}} \cdot \mathrm{i}\\
\\
&\Rightarrow \Re(\sqrt{z}) = \sqrt{\frac{\Re(z) + |z|}{2}} = \sqrt{\frac{\Re(z) + \sqrt{\Re(z)^{2} + \Im(z)^{2}}}{2}}\\
&\Rightarrow \Im(\sqrt{z}) = \operatorname{sign}(\Im(z)) \cdot \sqrt{\frac{-\Re(z) + |z|}{2}} = \frac{\Im(z)}{|\Im(z)|} \cdot \sqrt{\frac{-\Re(z) + \sqrt{\Re(z)^{2} + \Im(z)^{2}}}{2}}\\
\end{align*}
$$
If we use this for $ z^{2} $ with $ z = \Re(z) + \Im(x) \cdot \mathrm{i} $ follows:
$$
\begin{align*}
z^{2} &= \left( \Re(z) + \Im(z) \cdot \mathrm{i} \right)^{2} = \Re(z)^{2} + 2 \cdot \Re(z) \cdot \Im(z) \cdot \mathrm{i} - \Im(z)^{2}\\
&\Rightarrow \Re(z^{2}) = \Re(z)^{2} - \Im(z)^{2}\\
&\Rightarrow \Im(z^{2}) = 2 \cdot \Re(z) \cdot \Im(z)\\
\\
\sqrt{z^{2}} &= \sqrt{\frac{\Re(z^{2}) + |z^{2}|}{2}} + \operatorname{sign}(\Im(z^{2})) \cdot \sqrt{\frac{-\Re(z^{2}) + |z^{2}|}{2}} \cdot \mathrm{i}\\
\sqrt{z^{2}} &= \sqrt{\frac{\Re(z)^{2} - \Im(z)^{2} + \sqrt{\Re(z)^{4} - 2 \cdot \Re(z)^{2} \cdot \Im(z)^{2} + \Im(z)^{4} + 4 \cdot \Re(z)^{2} \cdot \Im(z)^{2}}}{2}}\\
&+ \frac{\Re(z) \cdot \Im(z)}{|\Re(z) \cdot \Im(z)|} \cdot \sqrt{\frac{-\Re(z)^{2} + \Im(z)^{2} + \sqrt{\Re(z)^{4} - 2 \cdot \Re(z)^{2} \cdot \Im(z)^{2} + \Im(z)^{4} + 4 \cdot \Re(z)^{2} \cdot \Im(z)^{2}}}{2}} \cdot \mathrm{i}\\
\end{align*}
$$
aka to simplify it with this might be really hard and/or no real simplification ...
(pls don't forget that x / |x| with x = 0 is here 0 'caused by sign...)
A: The function square root $\sqrt{z}$ is defined on the complex plane minus a curve connecting the points $z = 0$ and $z =\infty$. The curve is usually taken as the positive or negative half-axis of the real line. Therefore, the result $\sqrt{z^2}$ depends on the chosen branch. Moreovere, if you take a point on the curve, be careful which side of the curve cointains this point.
A simplification can be implemented, for instance, in Mathematica by a description of the fixed branch. Maybe the following example in Mathematica will clarify the role of the branch definition:
In[1]:= f[z_] = Sqrt[z^2 - 1] - Sqrt[z + 1] Sqrt[z - 1];
        f[I - 1.] // Simplify
Out[2]= 1.5723 - 2.54404 I

