Alternatively, I could use my functions, u and v, to evaluate the function, f.

So, writing z as x + iy, 2 + i means that x is 2 and

y is 1, the number that i gets multiplied with here.

And so, u(x,y) and I plug in x = 2 and y = 1.

So, it's u(2,1).

And since the function, u, is x squared minus y squared, it takes the first

parameter, that's everything squares it, and subtracts from now the second

parameter squared, so 2 squared minus 1 squared, that's 4- 1, that's 3.

And similarly, v(x,y) would be v(2,1).

And since v takes those two parameters, multiplies them and

multiplies the result by 2, I would get 2 times 2 times 1, which is 4.

So f(2 + i) can also be evaluated by taking u(2,1) + iv(2,1),

u(2,1 )was 3, v(2,1) was 4, so 3 + 4i, and that

is indeed the same result we obtained by just plugging in z after the function, f.

So these are two alternative ways of looking at the same function, and

they'll become helpful when we want to look at the function, u, and

the function, v, and their differentiability properties.

What do we already know about the function z squared?

Well we know this function is differentiable everywhere in C, in fact,

we found its derivative to be 2z for all z and C.