Math 226

Solutions to HW #4

October 14, 2003

p. 68, 1c, 2d, 3c, p. 73, 1d, p. 90, 8b, p. 94, 2b, 3a,b, 4, p. 104, 10, 11, 13

p. 68, 1c. We're given the function

As usual, this means so we could write this (and it might be clearer to write this) as

Anyway, if we let and then we have

and

Consequently, the Cauchy-Riemann equations, which say

would only be satisfied if

The second equation can only be satisfied if but under those circumstances, the first equation cannot be satisfied at all. Hence there is no point at which both these equation are satisfied. Thus, there is no point at which the function is differentiable.

2d. We consider the function

We want to verify first that the real and imaginary parts of satisfy the Cauchy-Riemann equations.

We have, writing

and we compute:

and

Now we can see that the Cauchy-Riemann equations are satisfied, and since these partial derivatives are continuous functions of and we know that is differentiable everywhere. Moreover, we know that

so

Now we can repeat this process and show that is itself differentiable.

We have where

so that

and

Again we see that the C-R equations are satisfied, and since these partial derivatives are everywhere continuous, is differentiable everywhere and

Notice this second derivative is just the negative of the original function We are not surprised, since we recognize that

3c. Now we're given the function

Denoting the real and imaginary parts of by and , as usual, we compute

and

Thus, the C-R equations are satisfied when

These equations are satisfied only at That is, only at But the partial derivatives exist everywhere and are continuous everywhere, so that's (more than) enough to meet the requirements of the theorem in section 21. So is differentiable at and the derivative is

p. 73, 1d. We're given the function

Now we have the real and imaginary parts of so we can hope to verify that the C-R equations hold. (We already see that the partial derivatives will be continuous, since the real and imaginary parts are sums and products of functions that we know have continuous partials of all orders).

Now we calculate

and

and

and

Now we see that and

p. 90, 8b. We want all values of such that

In other words, we want to calculate We know that

so

where runs through all the integers.

p. 94, 2b. As in the previous problem, we have

where runs through all the integers. More specifically,

where runs through all the integers.

3a. We have

On the other hand

3b. We have

On the other hand

4a. Suppose we choose the branch of the logarithm that says

where we require

Then we have

And

Thus, for this branch of the logarithm,

b. But if we take the branch of the logarithm that says

where we require

Then we have

And

Thus for this branch of the logarithm,

p. 104, 10a. We are told that

Therefore,

and so

b. Similarly, We are told that

for all and

Therefore,

and so

11a. Again, from the equality

we get

and so

But we also have

and the desired inequality is

which we can then rewrite as

or

or

But this we know is true for all

11b. In a similar way we are told that

so

and so

But also

so the desired inequality

becomes

or

or

which we know to be true for all

13. We have

so

Now we compute the partial derivatives:

Also

We see that the C-R equations are only satisfied when

But is never so the first equation becomes

which means

But at these values of so the second equation requires that and so Thus the C-R equations are satisfied at all the points on the real axis of the form

The partial derivatives of the real and imaginary parts of are continuous everywhere, so is differentiable at each of these points. However, for to be analytic at a point, it must not only have a derivative there, but it must have a derivative at every point in some neighborhood of that point. Since the above points are the only points where has a derivative, there is no neighborhood throughout which has a derivative, and so is analytic nowhere.