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Which of the following statements are true?

a. There exists an analytic function $f : \mathbb C → \mathbb C $ such that its real part is the function $e^x$, where $z = x + iy.$

b. There exists an analytic function $f : \mathbb C → \mathbb C $ such that $f(z) = z$ for all $z$ such that $|z| = 1$ and $f(z) = z^2$ for all $z$ such that $|z| = 2.$

c. There exists an analytic function $f : \mathbb C → \mathbb C $ such that $f(0) = 1, f(4i) = i$ and for all $z_j$ such that $1 < |z_j | < 3, j = 1, 2,$ we have $|f(z_1) − f(z_2)| ≤ |z_1 − z_2|^{\frac{\pi}{3}}$ .

My Try:-

(a) The real and imaginary part of an analytic functions are harmonic. That is $\frac{d^2e^x}{d^2 x}+\frac{d^2e^x}{d^2 y}=e^x \neq 0$. So, $f$ is not analytic. This is not the case. So. (a) must be false.

(b) Let $w=f(z)$, I think it is possible. $|z|=1$, $f(z)=z$. That is maps a $|z|=1$ to $|w|=1$ $|z|=2 \implies z=2e^i\theta \implies f(z)=4e^{i 2\theta}. $ which maps to a semicircle of radius $4$ and centre $0.$ Cant we join suitable sheets and paste with these two curve to form an analytic function.

(c) $\frac{|f(z_1) − f(z_2)|}{|z_1 − z_2|} ≤ |z_1 − z_2|^{\frac{\pi}{3}-1}$. So $z_2 \to z_1$. we get $f'(z)=0, \forall z:$ $1 < |z | < 3(\text{connected domain})$. hence $f(z)=k(k \text{is a constant} )$. So, there is an analytic function. $g(z)=f(z)-k$. Which is not possible. Since zeroes of analytic functions are isolated.

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  • $\begingroup$ In b), will $f(z)-z$ have isolated zeroes? $\endgroup$ – Gerry Myerson Jun 7 at 3:07
  • $\begingroup$ How do I check? $\endgroup$ – Math geek Jun 7 at 3:13
  • $\begingroup$ It is staring you in the face. If $f(z)=z$ for $|z|=1$, then the zeroes of $f(z)-z$ will incvlude ....? $\endgroup$ – Gerry Myerson Jun 7 at 3:15
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    $\begingroup$ okay. Thank you :) $\endgroup$ – Math geek Jun 7 at 4:09
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    $\begingroup$ @Mathgeek But the constant function $f(z) = 0$ is the only entire function which has non-isolated zeroes, by analytic continuation. $\endgroup$ – Ovi Jun 7 at 18:32

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