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Let $f(z)=\exp\left(\dfrac{z+1}{z-1}\right),\quad (z\in\mathbb{D}=\{z: |z|<1\})$

(a) Find the first three Taylor coefficients of f, that is, the constants $a_1$, $a_2$, $a_3\in\mathbb{C}$ such that $$f(z)=a_0+a_1z+a_2z^2+O(z^3)$$

(b) Find a finite Blaschke product $B$ such that $$B(z)=a_0+a_1z+a_2z^2+O(z^3)$$

My attemp: We have $f(z)=f(0)+\dfrac{f'(0)}{1!}z+\dfrac{f''(0)}{2!}z^2+O(z^3)$ and \begin{eqnarray*} f(0)&=&e^{-1}\\ f'(z)&=&\dfrac{-2}{(z-1)^2}\exp\left(\dfrac{z+1}{z-1}\right) \quad \Rightarrow f'(0)=-2e^{-1}\\ f''(z)&=& \dfrac{4}{(z-1)^3}\exp\left(\dfrac{z+1}{z-1}\right)+\dfrac{4}{(z-1)^4}\exp\left(\dfrac{z+1}{z-1}\right)\quad \Rightarrow f''(0)=0 \end{eqnarray*} So $$f(z)=f(0)+\dfrac{f'(0)}{1!}z+\dfrac{f''(0)}{2!}z^2+O(z^3)= e^{-1}-2e^{-1}z+0z^2+O(z^3)$$

$‎\underline{ For (b):}$ We know that a finite Blaschke product $B$ is the form $$B(z)=\prod_{n=1}^{N}\dfrac{|z_n|}{z_n}\dfrac{z_n-z}{1-\bar{z_n}z}$$ We should find a finite Blaschke product $B$ such that $$B(z)=a_0+a_1z+a_2z^2+O(z^3)=e^{-1}-2e^{-1}z+0z^2+O(z^3)$$ We have ‎ ‎$$‎B_0(z)=\tau_{a_0}(z)=‎\dfrac{a_0-z}{1-‎\overline{a_0}z‎}‎=‎\dfrac{e^{-1}-z}{1-‎\overline{e^{-1}}z‎}‎=‎\dfrac{1-ze}{e-z}‎\qquad (*)‎‎‎$$‎ ‎and ‎‎‎‎$‎B_{N+1}(z)=\tau_{a_0}(zB_N(z))‎$ So‎ $$B_1(z)=\tau_{a_0}(zB_0(z))\displaystyle\mathop{=}^{(*)}‎\dfrac{1-zB_0(z)}{e-zB_0(z)}‎‎=‎\dfrac{1-z‎\frac{1-ze}{e-z}‎}{e-z‎\frac{1-ze}{e-z}‎}‎=‎\dfrac{z^2e-2z+e}{z^2e-(1+e)z+e^2}‎$$ and $$B_2(z)=\tau_{a_0}(zB_1(z))\displaystyle\mathop{=}^{(*)}‎\dfrac{1-zB_1(z)}{e-zB_1(z)}‎=‎ ‎\dfrac{1-z\cdot ‎\frac{z^2e-2z+e}{z^2e-(1+e)z+e^2}}{e-z\cdot ‎\frac{z^2e-2z+e}{z^2e-(1+e)z+e^2}}$$ But I don't Know how I can continue?

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  • $\begingroup$ I believe your first and second derivatives are incorrect. In fact, the denominator should be $(z-1)$ instead of plus. Also, on the second derivative a $z$ factor should appear such that $f''(0)=0$. $\endgroup$ – Anthony Apr 9 '18 at 2:51
  • $\begingroup$ Yes, you're right. Thanks $\endgroup$ – Fair Apr 9 '18 at 13:18
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Take $N=3$ and take real $z_n$'s. Expand $\frac 1 {1-\bar {z_n}z} $ in a geometric series. Now you can see how to choose $z_1,z_2,z_3$.

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  • $\begingroup$ Could you explain it? $\endgroup$ – Fair Apr 3 '18 at 13:31

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