Minimum uncertainity I'm confused in finding the condition for minimum uncertainty, The author in the book I refer goes on saying that
$|g\rangle=c|f\rangle$
is the condition for minimum uncertainity
for some constant $c$
Where $|f\rangle=(\hat{A}-\langle A\rangle)\psi$
and  $|g\rangle=(\hat{B}-\langle B\rangle)\psi$
where $A$ and $B$ are any observables
This is acceptable due to the Schwarz inequaltity (since it becomes an equality when the angle between 2 vectors are zero), the author further says that this equality results if
${\operatorname{Re}\langle f|g\rangle}={0}$
I dont understand why this should be the condition , so if according to the author the angle should be zero, shouldn't
${\operatorname{Im}\langle f|g\rangle}$ be equal to ${0}$ ?
since for any varialbles $a$ and $b$ the argument of $z=a+ib$ can be zero only if the 
$\tan^{-1}\frac{b}{a}=0$ which implies  $b=0$ or $\operatorname{Im}{z}=0$ 
Kindly help me understand, any help is appreciated
 A: The generalized
uncertainty principle
relating two operators $A$ and $B$  is 
$$\sigma_A^2 \sigma_B^2 \geq {\left( \frac{1}{2i} \langle\left[ A,B\right] \rangle\right)} ^2$$
where $[A,B]=AB-BA$
is the
commutator of the
operators.
This relation was derived
using two inequalities.
The first is the Schwarz
inequality which, in bra-
ket notation is
$$\langle{f|f}\rangle\langle{g|g}\rangle \geq |\langle {f|g}\rangle |^2$$
The second inequality is
condition on a complex
number $z=x+iy$ :
$$|z|^2 \geq y^2$$
It’s interesting to see
what happens if we
require these two
relations to be equalities
rather than inequalities.
This will give us a
condition on the
functions f and g that
gives the minimum
uncertainty between
them.
In the case of the Schwarz
inequality, we want
$\langle{f|f}\rangle\langle{g|g}\rangle =|\langle{f|g}\rangle |^2$ .
To see what this implies,
we can examine the proof
of the Schwarz inequality.
To this end, we’ll
introduce the function,
$$|h\rangle =|g\rangle -\frac{\langle{f|g}\rangle}{\langle{f|f}\rangle} |f\rangle$$
since,
$\langle{h|h}\rangle \geq 0$ ,
by taking the product with itself,
$$\langle{h|h}\rangle =\langle{g|g}\rangle -\frac{\langle{f|g}\rangle}{\langle{f|f}\rangle} \langle{g|f}\rangle -\frac{\langle{g|f}\rangle}{\langle{f|f}\rangle} \langle{f|g}\rangle +\left( \frac{|\langle{f|g}\rangle |}{\langle{f|f}\rangle}\right)^2\langle{f|f}\rangle\\  
=\langle{g|g}\rangle -\frac{|\langle{f|g}\rangle |^2}{\langle{f|f}\rangle}\\  
\geq 0\\  
\implies \langle{f|f}\rangle \langle{g|g}\rangle \geq |\langle{f|g}\rangle |^2$$
In order for this inequality
to be replaced with an
equality, we would need
$|h\rangle =0$  .
Thus, from the definition, we have $|g\rangle=\frac{\langle{f|g}\rangle}{\langle{f|f}\rangle} |f\rangle$
That is, the Schwarz
inequality becomes an
equality if one function is
a scalar multiple of the
other:
$|g\rangle=c|f\rangle$
where c is, in general, a
complex scalar.
The second inequality is an equality if x=0
so that z is purely
imaginary. In our original
derivation of the
uncertainty principle, we
used $z=\langle{f|g}\rangle$ , so if we’re
requiring equality, we get $Re(z)=0$
[edit]:  
Here $z=\langle{f|g}\rangle$ is imaginary, because for minimum uncertainity we need the complex number to be puerly imaginary, as I stated early. And also $|g\rangle =c|f\rangle$ . Thus, $z=c\langle{f|f}\rangle$ . Since $\langle{f|f}\rangle$ is always real, c must be complex number.
A: I think I found out the answer to my own question
$|f\rangle$ and $|g\rangle$ are defined in Hilbert space which is an inner product space, and one of the properties of this space is 
$$\langle f|f \rangle \geq 0$$
So $\langle f|f \rangle$ should be real, 
In the derivation of the uncertainty principle we take 
$$(\text {Re}(z))^2+(\text {Im}(z))^2 \geq (\text {Im}(z))^2$$
where $z=\langle f|g\rangle$
The above inequality can be an equality when $\text {Re}(z)=0$, i.e $\text {Re}(\langle f|g\rangle)=0$
So we take $|f\rangle=c|g\rangle$ (due to Schwarz inequality), which gives $\text {Re}(c\langle f|f\rangle)=0$, since $\langle f|f \rangle$ is real (as stated above), $c$ must be complex
