<s>[INST] 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 $2\sqrt{3}$.
 [/INST] 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 $m=2\surd 3$.
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