umm good puzzle...

if we set α to zero then Is2 will be set to zero which means that now there will be an open circuit instead of the dependent current source.

this will change KCL since there wont be any current in the branch where Is2 was. this in turn will change the voltage drops on the resistors and eventually the currents in the circuit, including Ix.

E4 will still have no effect on the circuit (as i understand it, the voltage drop that was on Is2, is now the voltage drop on the open circuit)

im really not sure that this is correct..
ive always had difficulties with the "if we change x, will y change" kind of questions..

Spot on, and your reasoning is just fine. Technically, the current source is still there and what it is doing is the same thing it always did, namely produce whatever voltage is necessary to result in the programmed current flowing through it. In this case, that programmed current is zero and so the supply produces the same voltage that would appear between those nodes as if the source were removed entirely. The effect is the same, but it's handy to visualize how a source that was still there would have to act in order to behave as if it weren't.

 

Right on target! An existing current source, independently of its type, acts as an open circuit -regarding impedance- to the remaining elements; likewise, by the duality principle, an existing voltage source behaves as a short circuit. It is easier to grasp when the source is dependent because you can think of the controlling argument as a varying parameter ranging from, say, zero to whatever; if you "turn the knob" the magnitude of the voltage or current varies accordingly, but the internal equivalent impedance remains the same.