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

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Homework Statement:
.
Relevant Equations:
. False

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.

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Homework Statement:: .
Relevant Equations:: .

View attachment 294651

False

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.

• • docnet and S.G. Janssens
What can you say about the zeros of ##f## in the complex plane?

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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.

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

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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##.

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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!

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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.

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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.

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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

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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.

• docnet
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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 and PeroK
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
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so ##f=|z|^2-2zRe(z)## satisfies the Cauchy Riemann equations and so it's entire.
Would you believe it!

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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! • 