True or false? |z|^2 is an entire function

[docnet]

Homework Statement:

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Relevant Equations:

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False

The reasoning for answer:

The absolute value function is is not analytic wherever its argument equals zero. $f$ is not analytic at $z=0$ so it is not entire.

 

[PeroK]

Homework Statement:: .
Relevant Equations:: .

View attachment 294651

False

The reasoning for answer:

The absolute value function is is not analytic wherever its argument equals zero. $f$ is not analytic at $z=0$ so it is not entire.

That's a fairly significant logical fallacy!

In general, if $f(z) = f_1(z) + f_2(z)$, then even if $f_1$ and $f_2$ are not analytic, their sum may be. Intuitively, the non-analytic behaviour of $f_1$ and $f_2$ may cancel each other out in some way.

 

[S.G. Janssens]

What can you say about the zeros of $f$ in the complex plane?

 

[docnet]

That's a fairly significant logical fallacy!

In general, if $f(z) = f_1(z) + f_2(z)$, then even if $f_1$ and $f_2$ are not analytic, their sum may be. Intuitively, the non-analytic behaviour of $f_1$ and $f_2$ may cancel each other out in some way.

really? that is surprising to me.

 

[PeroK]

Take as an example: $$f(z) = z = Re(z) +iIm(z)$$ is the sum of two non-analytic functions!

 

[docnet]

What can you say about the zeros of $f$ in the complex plane?

the zeroes of $f$ are given by $$|z|^2=2zRe(z)$$
the left size is real valued, so the right side has to be real valued. The equation $x^2=2x^2$ has no solutions other than $x=0$, so the zero of $f$ is $z=0$.

 

[Infrared]

You have one thing a bit backwards: $z=0$ is actually the only point where $|z|^2$ is complex differentiable. You can verify this from the Cauchy Riemann equations.

And anyway, a function $f$ being (not) differentiable at a point doesn't mean the same is true for $f+g.$

For your function, you can also write it in real and imaginary parts and see whether the Cauchy-Riemann equations hold.

Edit: Apologies, while I was writing this, several posts appeared that i didn't notice!

 

[docnet]

Take as an example: $$f(z) = z = Re(z) +iIm(z)$$ is the sum of two non-analytic functions!

okay.. that makes sense

 

[docnet]

You have one thing a bit backwards: $z=0$ is actually the only point where $|z|^2$ is complex differentiable. You can verify this from the Cauchy Riemann equations.

And anyway, a function $f$ being (not) differentiable at a point doesn't mean the same is true for $f+g.$

For your function, you can also write it in real and imaginary parts and see whether the Cauchy-Riemann equations hold.

How do you tell the real and imaginary parts of $z$ in the absolute value function? this is what I wanted to do, compute the Cauchy Riemann equations, but the absolute value function confused me.

 

[PeroK]

okay.. that makes sense

You know what I'm going to say ... you did the same thing again! You jumped at an easy, one-line proof, that you didn't seriously question. You must be able to fully justify each step you take.

Developing your maths skills means developing the ability to find the example I gave in post #5.

 

[docnet]

How do you tell the real and imaginary parts of $z$ in the absolute value function? this is what I wanted to do, compute the Cauchy Riemann equations, but the absolute value function confused me.

Oh gosh.. I forgot that the absolute value function has a formal definition $|z|=\sqrt{z\bar{z}}$. I'm so sorry @PeroK

 

[PeroK]

How do you tell the real and imaginary parts of $z$ in the absolute value function? this is what I wanted to do, compute the Cauchy Riemann equations, but the absolute value function confused me.

The absolute value is real, it has no imaginary part.

 

[Infrared]

How do you tell the real and imaginary parts of in the value funciton? this is what I wanted to do, compute the Cauchy Riemann equations, but the absolute value funciton confused me.

You would write $z=x+iy$ (so $\text{Re}(z)=x$ and $|z|^2=x^2+y^2$).

 

[docnet]

$$f=|z|^2-2zRe(z)=x^2+y^2-2(x+iy)x$$
$$=(y^2-x^2)-i(2xy)$$
$$v_x=-2x=u_y$$
$$-v_y=-2y=u_x$$
so $f=|z|^2-2zRe(z)$ satisfies the Cauchy Riemann equations and so it's entire.

 

[PeroK]

so $f=|z|^2-2zRe(z)$ satisfies the Cauchy Riemann equations and so it's entire.

Would you believe it!

 

[docnet]

There goes my dreams for a Ph.D in maths, smashed to granules.

 

[PeroK]

There goes my dreams for a Ph.D in maths, smashed to granules.

You'd certainly bash out a PhD thesis quickly enough. Whether it would be defensible is another matter!