Complex Differentiation Homework Meme

Given a real value function f of a real variable so

there's nothing complex in here.

And a point x0 in the integral in which f is defined we say the function is

differentiable at x0 if the limit of the difference quotient exists.

And the difference quotient is f (x)- f (x ) divided by x- x 0.

If that limit as x goes to x 0 exists, we call the limit the derivative of f at x 0.

And denote it by f prime of x 0.

Let's quickly review what this means geometrically.

And then we take a second point x.

And we look at f ( x) and so this point that I'm now circling is the point

whose coordinates are x and f ( x).

Next we look at the numerator of our difference quotient f (x)- f ( x0).

F(x) is this value right here.

F(x0) is this value.

The difference between the two of them is this length.

So this is f(x)- f(x0).

And we divide that by x- x0.

X- x0 you can see in this length right here, so this thing right here.

F(x)- f(x0) divided by x- x0 is

rise over run for this red secant line that I drew.

In other words it gives us the slope of the red secant line.

It's the change in y divided by the change in x.

So this difference quotient f(x)- f(x0) divided by x- x0.

It's the slope of the secant line through the points (x0,

f(x0)) and (x,f(x)).

Next we need to let x approach x0.

And by doing that, the slope of the secant line will change.

As this point, x moves toward x0, the slope of

the red line that's the secant line will change, here's an example.

X moved closer to x0 and the line changed quite a bit.

As x moved closer and closer and closer to x0,

the secant line becomes a tangent line in the limit.

In the limit x and x0 have kind of overlapped.

And the slope of the secant line becomes the slope of the tangent line.

In other words, if this limit exists,

it is the slope of the tangent line to the graph of f at x0.

The derivative is thus the slope of the tangent line to the graph at x0.

The slope can be negative, it can be zero, it can be positive.

Here you see an example where the slope and that's the derivative is negative,

the tangent line has a negative slope, it's sloping downward.

In the middle example you see an example where the tangent line is horizontal.

It has a zero slope and therefore the derivative is zero at this point x0 where

the minus is tangent to the craft.

In this example here, the slope of the tangent line is positive and

therefore the derivative is positive.

And it becomes more positive as you move along this graph because

the line would get steeper and steeper.

The derivative does not have to always do this however.

Here's an example where the graph has some kind of corner.

So if this is my point x0 f(x0) and

that word to approach this point x0 with x values from the left.

And these secant lines all look a little bit like this and

definitely have a positive slope even the limit of slope is positive.

However, if I approach this point from the right.

My secant lines have a negative slope and approach a limit that's negative in slope.

And I can not find a one tangent line to the graph of f at this point x0.

There is no tangent line and so

there's no unique slope which means this function does not have a derivative at x0.

So in this example f prime of x0 does not exist.

The derivative is defined in a completely analogous way to the real value case.

We say a complex valued function f of complex value is complex differentiable.

And typically we don't really say complex differentiable.

We say differentiable if it is clear from the context that we

mean complex differentiable.

So a function is differentiable at a point z0 which must be on the domain of f,

obviously.

If the limit of the difference quotient exists we

look at f(z)- f(z0) divided by z- z0 and let z approach z0.

Now notice, though if you pick a point z0 in the complex plane.

There are lots of ways of approaching that point.

It's not just along the real axis like we have to look at in the real value case.

You could also come in the complex direction from the imaginary axis.

You could come at a 45 degree angle.

Or you could even approach in spirals.

There's lots of different ways of having z approach z0.

And so for this limit to exist it's a little harder to check

than in the real case.

If this limit exists it's denoted f prime of z0.

Sometimes we also use the notation df dz as a derivative

of f with respect to z at z0, or just d dz of f(z) at z = z0

Let's look at some examples.

Suppose f (z) is a constant function, so no matter what z you plug in,

you get the same constant value out of it.

If you prefer, you can just think about a specific constant like ten right here.

Take a z0 in c and look at this difference quotient,

what is f(z)- f(z0) divided by z- z0?

F(z) is this constant c, but f(z0) is also equal c because f is a constant function.

Divided by z- z 0.

The numerator evaluates to 0, the denominator's non zero.

So the quotient's equal to 0.

And as z approaches z0 it doesn't matter how z approaches z0,

this quotient is always zero and clearly has the limit zero.

Therefore the derivative of f is equal to zero for all points In the complex plan.

Sometimes instead of writing the difference quotient as f(z)-

f(z0) divided by z- z0.

We write z as z0 + h.

Where h is a complex number.

And the difference quotient, then would become instead of writing z here,

we're going to write z0 + h, so f(z0 + h)- f(z0).

And z- z0 then becomes, if we write z as z0 + h,

and subtract from that z0, what's left over is h.

So this is the exact same quotient, just written in a different way.

In terms of the difference h, rather than in terms of c.

So the difference quotient is f(z0 + h)- f(z0) divided by h, or often

we just write f(z + h)- f(z) over h when we get tired of writing z0 all the time.

And we'll take the limit as h approaches 0, because when z and z0 get closer and

closer to each other, it means that the difference z- z0, which is h, goes to 0.

So here's another example, let's look at the function f(z) = z.

So the function that assigns to each point just the same point.

So what is f(z0 + h)- f(z0) divided by h?

Well f(z0 + h) is just z 0 + h, which you see right here.

F(z0) is simply z0.

F just returns exactly the argument that you passed into it.

We divide by h, and you see in the numerator the z

0's cancel out with each other, and you're left with h over h.

H over h is 1.

No matter what the value of h is.

And as h approaches zero, it remains one.

No matter how H approaches zero.

Which means this limit exists.

And that means the derivative of f is equal to one for

all of Z in the complex claim.

So this is differential everywhere with derivative one.

The difference quotient then becomes f(z0) plus h, and

the function f just takes its input, in this case, z0 plus h, and squares it.

So that becomes (z0 + h) quantity squared- f(z0), which is z0 squared, divided by h.

This term, (z0 + h) quantity squared multiplies through

to z0 squared + 2 times z0 times h + h squared.

And then we subtract z 0 squared from this expression.

So we have a z 0 squared here and we subtract one here.

Which means the numerator simplifies to 2 z 0 h + h squared.

Next we can cancel an h out of the numerator, an h out of the denominator.

So, we are left with 2z0 plus h.

And as h goes to zero, this plus h term becomes less and less important so

that in the limit we are left with two times zo.

So the limit exists and

therefore f prime of z is two z for all z in the complex plain.

Let's look at another example f(z) = z to the n.

So, the function takes an argument and raises it to the nth power.

Therefore, the difference quotient becomes f of z0 plus h which

z 0 plus h to the nth power minus f(z0), which is just z0 to the n divided by h.

Now how do you multiply add z0 + h to the nth power?

We can do that using the binomial theorem.

But, we're actually only entered in the first few terms.

The binomial theorem says that this is equal to the sum

of n choose k times

z 0 to the k times h to the n minus k.

And this runs from k to zero to n.

You don't really need to know this binomial theory.

All you need to know is that when you multiply through, you have different

powers of z zeroes, and different powers of h which always add up to n.

And the binomial coefficients tell you how many of those terms you have.

So for example you'll have a z0 to the nth power and no h term.

Then you have a Z0 to the n minus 1st power and an h term.

Then a Z0 to the n minus second power and then h squared terms,

that would be is Z0 to the n minus third in h cube and so

forth, all the way to Z0 to the no power times h to n.

And the number of times these terms occur are given by the binomial coefficients.

Z0 to the n occurs once, h times z0 to the n- 1 occurs n times, which is n choose 1.

This coefficient here is n choose 2, and so forth.

None of these really matter for the limit.

In the limit the only terms that are really important are these

first two terms.

We're left with terms that all have an h or higher power of h in them.

The only one that has a power one of h is this first term n times h

times z0 to the n minus one.

So if we cancel out an h of the numerator and

denominator, this term has no h's left in it.

It becomes n times z0 to the n minus one.

All the other terms just lose one power of their h.

And have all these remaining terms left in here.

So as h approaches 0, we have some number here that gets multiplied by h, which

becomes smaller and smaller and smaller, and in the end becomes negligible.

In other words the limit of this expression is n times z0 to the n- 1.

Therefore the function z to the n is differentiable in the complex

sense with derivative n times z to the n minus one for all z and c.

Now that we have taken a few derivatives it would be great to understand how we can

combine functions whose derivatives we already know.

So let f and g be differentiable functions and differentiable at z, and

suppose h is another function that's differentiable at f(z).

We want to look at the composition that's why we want to have

h differentiable at f(z).

And suppose c is a complex constant.

The product is a little bit more complicated, it's not sufficient to

just simply differentiate each function, but there's a product rule.

You differentiate f and multiply it with g.

Next, you don't differentiate f and multiply that by the derivative of g.

There's also a rule for the derivative of the quotient f over g.

Obviously you can't do that if g(z) is 0.

So you want to require here that g(z) is not 0.

But if it is not 0 the quotient rule is the same as it is in calculus.

You find the derivative of the quotient by squaring the denominator,

writing down the denominator again times the derivative of the numerator,

minus this time the numerator times the derivative of the denominator.

So you almost see the product rule in the numerator except for that minus sign.

And so you need to remember which term comes first, which term is

the one with the minus sign and which term is the one without the minus sign.

The way I remember that is I square the denominator and

write it down right away again and then times the derivative of the numerator,

therefore I know which one is the term that comes first.

Rather than showing you all these proofs which are the same as the proofs in

calculus, I'm going to show you a number of examples.

These are important rules to practice.

If you are not so familiar with these rules from calculus anymore,

I highly recommend Let you do some practice.

You could, for example, practice with all the problems I'm about to present but by

trying to find the derivatives by yourself first before looking at my solutions

Let's start with a function f(z) = 5z cubed + 2z squared- z + 7.

So we have a sum of a bunch of functions.

Difference is the same as the sum,

you can therefore differentiate these four functions individually.

So what are all these derivatives?

The derivative of 5 times z cubed is a constant times z cubed.

Constants can go to the side, 5, times the derivative of z cubed.

We just found the derivative of z to the n.

That was n times z to the n- 1.

In other words, the derivative of z cubed is 3 times z squared.

So, the derivative of 5 z cubed is 5 times 3 z squared.

The derivative of z which is either the one that we found earlier,

which was 1, or it would also apply this as the derivative of z to the first power

which is 1 times z to the zeroth power, which is also 1.

And 7 is simply a constant whose derivative is 0.

Therefore, I didn't write down the plus 0 part.

Now if you wanted to simplify this a bit, this is 15z squared + 4z- 1.

Next, let's look at 1 over z.

We can use the quotient rule for that one.

Even though there's also easier ways of finding this derivative, but

that's just fine by using the quotient rule.

So we square the denominators, z squared, we write down the denominator again,

that's z, times the derivative of the numerator, which is 0.

The numerators are constant.

Its derivative is 0.

Minus the numerator, 1, times the derivative of the denominator.

The derivative of z is 1.

The numerator there for simplifies to 1 with a negative sign in front of it.

The denominator is z squared.

So the derivative of 1 over z is -1 over z squared.

A different way of finding this derivative would have been to look at this function

as the function f of z = z to the power -1.

And realizing that this rule for

the derivative of z to the n also holds for negative exponents.

And so the derivative is therefore -1 times z to the n -1.

So -1 -1, which is -z to the -2, and

if you rewrite that, that is -1 divided by z squared.

Let's look at f(z) = (z squared- 1) to the nth power.

So here we have an inside function and

an another function on the outside to the power n.

Is h prime of f of z times f prime of z.

So we need to find the derivative of the outside function,

which is the function that raises its argument to the nth power.

Well, the derivative of that is n times the argument to the n-1 power.

And then multiply that by the derivative of the inside.

So z squared- 1, well the derivative of z squared -1 is 2z.

Therefore the derivative in this case is n times

z squared minus 1 to the n minus 1st power times 2z.

Let's look at another equation.

f(z) = (z squared- 1) (3z + 4), which is the product of two functions.

Of course we could multiply through and then differentiate term by term.

But I want to show you a use of the product rule here.

The product rule said that we need to take the derivative of the first function

which in this case is 2Z, the derivative of z squared- 1.

And multiplying with the second function, 3z + 4 plus the first function,

z squared- 1 times the derivative of the second function, but the derivative of

3z + 4 is simply 3 because 4 is a constant and its derivative is 0.

Finally, let's look at z divided by z squared + 1, we can.

Let's use the quotient rule which means we have to square the denominator z

squared + 1, 1 is z squared.

Write down the denominator again and

multiply with a derivative of the numerator.

But the numerator is z, its derivative is 1.

So I didn't even write down the times one right here.

And if we wanted to simplify that, we have a (z squared + 1)- 2 z squared, so

the numerator becomes 1- z squared, and

the denominator is the same as it was before.

I reordered as (1 + z squared) quantity squared just to make it look nice.

Let's use the f (z + h)- f(z) over h form of the difference quotient.

And let's write h as hx, the real part of h, + i h y, the imaginary part of h.

Then what is f of z + h?

z + h would be x + h x + i y + h y.

And if we only look at the real part of that, that real part is x + hx.

We subtract from that f(z), f(z), this is x.

We see that there is an x here and we subtract x here, so that cancels out.

The numerator becomes simply hx.

The denominator is h, hx was the real part of h.

So the difference quotient is simply the real part of h divided by h.

What happens as h goes to 0?

As we noted earlier, there are lots of different ways of approaching

a point, h could go to 0 along the x axis.

Or it could go to 0 along the imaginary axis, or along all kinds of other ways.

But let's check out for

starters what happens when h approaches along the real axis.

So h in other words is real elements of

the form h x+ IX0 does not have an imaginary part.

In other words the real part of h which is

hx is the same as h itself because there is no imaginary part.

In other words in this quotient right here the numerator is equal to

the denominator which means the whole fraction evaluates to 1 and

as h goes to 0 just along the real axis the limit is 1.

So then it exists as h approaches 0 along the real axis.

So again in this fractional real part of h over h the numerator is 0.

So the whole fraction evaluates to 0.

And in the limit as h approaches 0 along that imaginary axis, the limit is still 0.

That's a different limit than we obtained coming along the real axis.

Следи за мной, - холодно парировал Стратмор. - А как же Сьюзан? - Хейл запнулся.  - Если вы позвоните, она умрет. Стратмора это не поколебало.

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