thevenin
In a practice test we are asked to find the Thevenin equivalent by finding the Thevenin voltage and Norton current within the following circuit.
We were given the following answer for Vth:
I feel like I'm making a simple mistake but I'm coming up with
Vth= 
which is equal to 
as I see he wrote down. So why does it wind up as 7+7j?
Redraw the circuit as follows to see why \$V_X\$ is across the 60 ohm resistor, the controlled current source and the output terminals:
It is often helpful to remember that two points connected by an ideal wire are the same circuit node which can be seen by 'shrinking' the wire to zero length as I've done above.
If you think about it that VCVS is not dependent. It depends on some parameters that are not part of the circuit you are analyzing, so you can turn it off (better than saying "short it" in my opinion.
You can also see this from another point of view: if you include the \$V^+V^-\$ part of the circuit and connect a generator to these terminals, then you should turn it off, then \$A(V^+-V^-)=0\$. Or you can just say that if you only have dependent sources in a circuit, then the "all nodes are at 0V" solution works, and some backend math tells us that in a nice network the solution is uniqe if it exists, so also if you don't include the generator on the op amp input terminals you can say well the output must be zero.
Understood this the solution is trivial and it's precisley what your book reports.
Best Answer
Looks like you are both correct, although you are more precise. The "7's" in the result 7 + 7j are an approximation of 5*sqrt(2) which somewhat more precisely = 7.07107, so I would prefer the solution given in rectangular form be 7.1 + 7.1j since 7.07.... rounds closer to 7.1 than 7.0