Okay, let me explain. What you have here is a complex number. Right now we have to put it in form #a+bi#. In that form we call #a# the *real part* and #b# the *imaginary part*. This because #b# is multiplied by #i#, which is nowhere on the real line. This makes things much easier. You will often see the notation of #Im(z)# as a function that gives you back the imaginary part and #Re(z)# as the function that give you the real part (ex. #Im(4+3i)=3#, #Re(4+3i)=4#). I'm going yo use that notation, so you can start gettin used to it.

Let's make #4+2i-(a+4i)# into #a+bi# form.

#4+2i-(a+4i)=4+2i-a-4i=4-a+2i-4i=(4-a)+(2-4)i=(4-a)-2i#

So we know that #(4-a)-2i=9-2i#. We can use the #Re# function here now, to solve for #a#.

#Re((4-a)-2i)=Re(9-2i)#

#4-a=9#

And so, we finally claim our victory:

#4=9+a#

#a=-5#