Finding modulus of $\sqrt{6} - \sqrt{6}\,i$ I found the real part $=\sqrt{6}$.
But I don't know how to find imaginary part. I thought it was whatever part of the function that involved $i$, with the $i$ removed? Therefore the imaginary part would be $-\sqrt{6}$.
Meaning the modulus is equal to 
\begin{align}
 \sqrt{ (\sqrt{6})^2 + (-\sqrt{6})^2}            = \sqrt{12}.
\end{align}
The answer was $2\sqrt{3}$.
 A: Yes, you are right:)
maybe it is of some help if you add a geometric interpretation of this, so the nex time it is easier for you to find modulus or real/imaginary parts.
A complex number $z=x+iy$ it can be viewed as a point of the complex plane $\mathbb{C}$ which is isomorphic to $\mathbb{R}^2$. So for each complex number $z$ there is a unique point $(a,b) \in \mathbb{R}^2$ who represent the number. Usually we use the isomorphism $\varphi: \mathbb{C} \to \mathbb{R}^2$, $\varphi(z)=(x,y)$. So in fact the imaginary part can be viewed as the y-coordinate of the number and the real part represents the x-coordinate in the plane. With this representation the module is the Euclidean distance from the origin. Just to be exhaustive at this point is obvious to introduce the polar representation of a complex number which is in fact another isomorphism between $\mathbb{C}$ and $\mathbb{R}^2$, provided that the angle is in $[0,2\pi[$, which uses modulus and angle of a complex number to identify it univocally:)
summing up, checking what are the parts "involved with the i" is correct for finding the imaginary part. Remember that each part (imaginary and real) are REAL numbers (the map goes in the real plane) 
A: To find the modulus of a complex number or a number in the form $a+b\sqrt{c}$, you multiply it by its conjucate.  This replaces $\sqrt{x}$ by $-\sqrt{x}$, where $i=\sqrt{-1}$.  
So conjucate of $a+bi$ is $a-bi$   ,  whence  $m^2 = (a+bi)(a-bi) = a^2 - (bi)^2 = a^2+b^2$
For the example in OP, the values of $a^2$ and $b^2$ are $\surd{6}$, gives $m^2=6+6=12$, whence $m=2\surd 3$.
