I Find f(z) given f(x, y) = u(x, y) + iv(x,y)

[PLAGUE]

TL;DR Summary

Why we replace y with 0 and why this is equivalent to replacing x with (z + z')/2 and y with (z - z')/2i?

I am given, $$u = e^{-x} (x sin y - y cos y)$$ and asked to find v such that, $$f(z) = u + iv$$. My book solves these problems and the answer is, $$v = e^{-x} (ysiny + x cos y) + c$$. I understand how it is done, using Cauchy-Riemann equations.

Then, the book asks to find f(z). When doing that, it says, $$f(z) = f(x+iy) = u(x, y) + iv(x, y)$$
Putting y = 0, $$f(x) = u(x, 0) + iv(x, 0)$$
Replacing x by z, $$f(z) = u(z, 0) + iv(z, 0)$$
The answer: $$f(z) = ize^{-z}$$

Why is replacing y with 0 is valid here?

The book solves this method in another method. It replaces x with $$\frac{z + \overline z}{2}$$ and y with $$\frac{z - \overline z}{2i}$$. Then, it simplifies and the answer is still the same, $$i z e^{-z}$$.

Why is this method equivalent to the previous one?
N.B. I know that you get the real part of z by $$\frac{z + \overline z}{2}$$ and the imaginary part by $$\frac{z - \overline z}{2i}.$$

I am giving the full answer here. They are taken from
by Murray SPIEGEL (Author)

 

[PeroK]

I don't see how it's valid to set $y = 0$. For example, it doesn't work here:
$$f(z) = x + 2iy \ \implies \ f(x) = x \ \implies \ f(z) = z$$The second method is valid, because that's just a regular substitution of $x, y$ by equivalent complex expressions involving $z$. E.g:
$$f(z) = x + 2iy \ \implies \ f(z) = \frac{z + \bar z}{2} + z - \bar z = \frac 3 2 z - \frac 1 2 \bar z = x + 2iy$$And that will always work.

 

[PLAGUE]

I don't see how it's valid to set $y = 0$. For example, it doesn't work here:
$$f(z) = x + 2iy \ \implies \ f(x) = x \ \implies \ f(z) = z$$The second method is valid, because that's just a regular substitution of $x, y$ by equivalent complex expressions involving $z$. E.g:
$$f(z) = x + 2iy \ \implies \ f(z) = \frac{z + \bar z}{2} + z - \bar z = \frac 3 2 z - \frac 1 2 \bar z = x + 2iy$$And that will always work.

Perhaps I am missing something. I am giving the full answer here. They are taken from

Schaum's Outline of Complex Variables, 2ed: Second Edition (Schaum's Outlines)​

by Murray SPIEGEL (Author)

 

[PeroK]

Perhaps I am missing something. I am giving the full answer here. They are taken from

Schaum's Outline of Complex Variables, 2ed: Second Edition (Schaum's Outlines)​

by Murray SPIEGEL (Author)View attachment 363833View attachment 363834

These are analytic functions. That explains it.

 

[PLAGUE]

The book is wrong. Method 1 is, in general, nonsense.

I was said that setting y equal to 0 is equivalent to setting $z=\overline z$ as done here. (see the lower half of the page) But it also doesn't make sense to me why you would set $z = \overline z$

 

[PeroK]

I was said that setting y equal to 0 is equivalent to setting $z=\overline z$ as done here. (see the lower half of the page) But it also doesn't make sense to me why you would set $z = \overline z$

Okay, I see now that you are dealing with analytic functions. Wherever $x$ appears, it always appears with $+iy$. The function has the same format in $x$ as it does in $z$. So, if you find $f(x,0)$, then you know $f(z)$ must look just the same.

Sorry, I hadn't looked at complex analysis in a while. I forgot how restrictive the analytic condition is.

 

[PLAGUE]

Okay, I see now that you are dealing with analytic functions. Wherever $x$ appears, it always appears with $+iy$. The function has the same format in $x$ as it does in $z$. So, if you find $f(x,0)$, then you know $f(z)$ must look just the same.

Sorry, I hadn't looked at complex analysis in a while. I forgot how restrictive the analytic condition is.

Can you suggest some study materials?

 

[PeroK]

Can you suggest some study materials?

Sorry, I haven't studied complex analysis properly since 1983!

 

[FactChecker]

TL;DR Summary: Why we replace y with 0 and why this is equivalent to replacing x with (z + z')/2 and y with (z - z')/2i?

I am given, $$u = e^{-x} (x sin y - y cos y)$$ and asked to find v such that, $$f(z) = u + iv$$

is analytic. Don't leave that part out. It is critical.

My book solves these problems and the answer is, $$v = e^{-x} (ysiny + x cos y) + c$$. I understand how it is done, using Cauchy-Riemann equations.

Then, the book asks to find f(z). When doing that, it says, $$f(z) = f(x+iy) = u(x, y) + iv(x, y)$$
Putting y = 0, $$f(x) = u(x, 0) + iv(x, 0)$$
Replacing x by z, $$f(z) = u(z, 0) + iv(z, 0)$$
The answer: $$f(z) = ize^{-z}$$

Why is replacing y with 0 is valid here?

Two analytic functions that are equal on the real line must be identical. So we only need to show that $$f(z) = ize^{-z}$$ on the real line.

The book solves this method in another method. It replaces x with $$\frac{z + \overline z}{2}$$ and y with $$\frac{z - \overline z}{2i}$$. Then, it simplifies and the answer is still the same, $$i z e^{-z}$$.

Why is this method equivalent to the previous one?

Without seeing the details, it is hard to say. Maybe it ends up with the same answer but uses different facts about analytic functions. I wouldn't call that an "equivalent" method.

N.B. I know that you get the real part of z by $$\frac{z + \overline z}{2}$$ and the imaginary part by $$\frac{z - \overline z}{2i}.$$

I am giving the full answer here. They are taken from
by Murray SPIEGEL (Author)

This link just takes me to the main amazon page. Maybe my browser (Firefox) is not working right.

This doesn't say enough to answer your question. You would have to go through the "tedious labor" to know the answer.