In this problem. We're looking at the third ah, function in the separated Schrodinger's equation. Uh, which will look like this sometimes with are as a function of little our theater as a function of little data. Oops, that's not right. I meant to write, uh, capital fee here, and capital fee is a function of little feet. So here we're just looking at a function representing fee as a mouthful angle. And what we want to show is that if we have this function and we give it an argument little fee, this is always going to be equal to, ah, the function. If we had give it feet plus two pi. So basically, what we're trying to show is that capital fee is a function that is two pi periodic. Um, So how do we do this? We're gonna have to remember Oilers formula, which is that e to the I theta is equal to co sign of data, plus I sine of data. So if we use this starting with what I'll call the left hand side over here, the original argument Well, we're gonna have that, um e so the M fee or I am fee is gonna be equal to co sign times m fee, Plus I sine of EMC. So this is our left hand side. So all we need to do now is sure that the right hand side is actually equal to this. Um, so this is our left hand side. Okay, So what about our right hand side if you go ahead and plug in feet plus two pi instead of just fee on what we're gonna get, um, I'll just write the whole thing out here is that feet plus two pi. This is going to be equal to e to the I i m times fee plus two pi and then this using oilers. Um, well, actually, let's expand this. First, we can say that using the properties of exponents, we have I and fee times e to the I to pi. OK, so using oilers on both of these terms here, what we're gonna get is actually co sign of m V plus I sine off and see. Um and this could be multiplied by the second term. Um, which is going to be co sign of too high? Oh, I forgotten m here. I've forgotten m here because this M is actually out here. It's one affected out. I had to Ah, distribute. Okay, so we have actually two pi em plus i sine to hi m. And one thing you'll notice is, um, any for any value of em. The critical realization here is that m has to be an integer. So this is just saying that M must be in the set of all integers from zero to plus minus. One plus one is two and so on. So this must be true. That is a defining property of our magnetic quantum number. M. So that is something we need to realize here. And if this is true, then for any value of m ah co sign of two pi times, some integer is always going to be equal toe one. It's always going to be equal to one. And similarly, the sign of two pi times. Any riel integer here is going to be zero. If you just think about how a sign looks anytime you hit some multiple of two pi, it's going to be zero. So that this leaves us with the right hand side, um, of this whole thing, which is the same is our left hand side times one. So that, um, basically concludes our work here, we can say that the left hand side is equal to the right hand side, which is saying that, um, big fee, given a function of fee, is the same as big feat, given a function of feet plus two pi.