Now given any particular f of zeta,

we can substitute zeta by x plus iy,

and we have a function of x and y with real and imaginary parts.

So, let's take a look at this example of

quadratic function: f of zeta equals zeta squared,

equals parenthesis x plus iy parenthesis square.

If we expand this,

then it will be x square minus y square,

which is the real part and plus two ixy,

which is the imaginary part.

We can rewrite that in this kind of notation where u of xy is the real part function,

v of xy is the imaginary part function,

and separate them out like here.

So, from any complex function f of zeta, two new functions,

u of xy and v of xy can be derived as you can see here,

and in the later slides,

we will show you that these functions have very interesting properties.

Now, before going that,

we will claim that for any ordinary function,

the functions u and v automatically satisfy the relations

written down here which is round u over round x equals round B round y,

and round v or round x equals minus round u over round y.

From this relationship, we can we can derive that they

fulfill the Poisson's equation for neutral object in two-dimensional space.

So, that's why you see here,

round square u over round x squared plus round square u over round y square equals zero,

and the same applies for v as well.

Okay? So, we're going to prove what we just claimed

for any ordinary function using the mathematical manipulation.

So, you can see f of x plus iy equals u of xy plus i times v of xy,

where we have two functions instead of one.

So, we're going to separate out each term and use the chain rule to prove this.

But before doing that,

let me ask Melodie, what is chain rule?

Okay. So, we have the chain rule actually written right here,

and what the chain rule implies is that you can break derivatives up,

so they become easier by replacing some component with another variable.

So, in this case, we have the derivative x taken in terms of some component z,

and then if you multiply that by the derivative of z in terms of x,

then you actually get the function in term taken into derivative of z,

and sometimes it's easier to break the derivative off in this manner.

Exactly. So, let's take a look at here.

Look at this equation round f over around x equals round f over around z,

so z is another parameter that we introduced to make it easier to solve.

Then we need to put round z over around x to make it equivalent.

Now, looking at this equation,

look at this, round z over around x is one.

Why is it one?

Because, z, we defined is equal to x plus iy.

Right? So, if I do partial derivative of z with respect to x as you can see here,

this one becomes one.

Right? So, because this is one,

we can ignore this and we only have this.

So round f over round x is really equal to round f over around z.

Using the same logic,

we can apply it to round f or round y,

and in this case round z over around y will be i,

because it's the imaginary part.

So, it is i times round f or round z.

Now, knowing this, now we can play with this round u over x plus i round v over round x,

this is round f over around z.

Why? Because round f over round x is equal to round f over round z.

So, we replaced round f over around x by round f over round z.

Right? Here too, round u over around y plus i round u over around y,

we replace round f over round y by i round f or round z.

Now with this two equations,

if I do a subtraction or an addition, what happens is,

subtraction over addition, by multiplying i to this equation,

because then these two becomes equivalent,

then we have real parts and imaginary part and by corresponding each part to each other,

we come up with these two equations.

Which is the same equations as we found in our slides,

just one slide before this.

Okay? Now.

So, thus starting with any ordinary function, right?

We can arrive at two functions,

U of x, y and V of x,

y, which are both solutions of Laplace's equation in two dimensions.

We just proved that right? We just proved that.

So, each function represents

a possible electrostatic potential because they fulfill Poisson's equation.

We can pick any function f over zeta,

which represents two electric field problems.

We can write down as many solutions as possible by

just making a functions and find a problem that goes with each solution.

So then we will come back to this quadratic function,

where we had real part x squared minus y squared,

and the imaginary part 2xy.

If we plot that in this graph,

if you plot that in this graph,

2xy will be this one or this one,

the dotted function, and x squared minus y squared will be this one,

the line that is connected.