Question

c. Evaluate ,f(z) dz with า the circle of radius 1 centered at the origin and traveled once counterclockwise ˊ们: (1-2 For real twith-1 < t < 1 and +12)-1 Explain why f(:)) has an expansion of the form in C , let f(z) be defined by fG)- a. b. Compute Uo(t), Ui(t), and Uz(t) in terms of t. c. Recalling that t is a real number smaller than 1 in absolute value, find the radius of convergence of this power series. (Hint: where are the singularities of fr(a) as a function of :?) # s, let f(s)

Answer 1

a) We solve the equation $\begin{align*}z^2-2tz+1=0\end{align*}$ for variable $\begin{align*}z\end{align*}$ (via quadratic formula), to get

$\begin{align*}z=t\pm \sqrt{t^2-1}\end{align*}$

Let $\begin{align*}\omega_+=t+\sqrt{t^2-1}\end{align*}$ and $\begin{align*}\omega_-=t-\sqrt{t^2-1}\end{align*}$ ; since $\begin{align*}-1<t<1\end{align*}$ , we have $\begin{align*}t^2<1\end{align*}$ , so that $\begin{align*}\omega_+,\omega_-\end{align*}$ are complex (and not real). Because $\begin{align*}z^2-2tz+1=(z-w_+)(z-w_-)\end{align*}$ , we find

$\begin{align*}(w_+-w_-)f_t(z)&={\frac {(w_+-w_-)}{(1-2tz+z^2)}}\\ &={\frac {(w_+-w_-)}{(\omega_+-z)(\omega_--z)}}\\ &={\frac 1{(\omega_--z)}}-{\frac 1{(\omega_+-z)}}\\ &={\frac {\frac 1{w_-}}{(1-{\frac z{\omega_-}})}}-{\frac {\frac 1{w_+}}{(1-{\frac z{\omega_+}})}}\\ &={\frac 1{w_-}}\begin{pmatrix}1-{\frac z{\omega_-}}\end{pmatrix}^{-1}-{\frac 1{w_+}}\begin{pmatrix}1-{\frac z{\omega_+}}\end{pmatrix}^{-1}\end{align*}$

Recall that $(1-x)^{-1}$ has a power series representation:

$(1-x)^{-1}=\sum_{n=0}^\infty x^n$

Therefore, the above shows

$\begin{align*}(w_+-w_-)f_t(z)&={\frac 1{w_-}}\begin{pmatrix}1-{\frac z{\omega_-}}\end{pmatrix}^{-1}-{\frac 1{w_+}}\begin{pmatrix}1-{\frac z{\omega_+}}\end{pmatrix}^{-1}\\ &={\frac 1{w_-}}\sum_{n=0}^\infty \begin{pmatrix}{\frac z{\omega_-}}\end{pmatrix}^n-{\frac 1{w_+}}\sum_{n=0}^\infty\begin{pmatrix}{\frac z{\omega_+}}\end{pmatrix}^n\\ &=\sum_{n=0}^\infty w_-^{-n-1}z^n-\sum_{n=0}^\infty w_+^{-n-1}z^n\\ &=\sum_{n=0}^\infty (w_-^{-n-1}-w_+^{-n-1})z^n\\ \Rightarrow\hspace{2cm}f_t(z)&=\sum_{n=0}^\infty (w_-^{-n-1}-w_+^{-n-1})(w_+-w_-)^{-1}z^n\end{align*}$

Thus, letting

$\begin{align*}U_n(t)&=(w_-^{-n-1}-w_+^{-n-1})(w_+-w_-)^{-1}\\ &=\begin{pmatrix}(t-\sqrt{t^2-1})^{-n-1}-(t+\sqrt{t^2-1})^{-n-1}\end{pmatrix}\begin{pmatrix}(t+\sqrt{t^2-1})-(t-\sqrt{t^2-1})\end{pmatrix}^{-1}\\ &=\begin{pmatrix}(t-\sqrt{t^2-1})^{-n-1}-(t+\sqrt{t^2-1})^{-n-1}\end{pmatrix}\begin{pmatrix}2\sqrt{t^2-1}\end{pmatrix}^{-1}\end{align*}$

we obtain the power series representation

$\begin{align*}f_t(z)&=\sum_{n=0}^\infty U_n(t)z^n\end{align*}$

b) Since

$\begin{align*}U_n(t)&=\begin{pmatrix}(t-\sqrt{t^2-1})^{-n-1}-(t+\sqrt{t^2-1})^{-n-1}\end{pmatrix}\begin{pmatrix}2\sqrt{t^2-1}\end{pmatrix}^{-1}\end{align*}$

we get

$\begin{align*}U_0(t)&=\begin{pmatrix}(t-\sqrt{t^2-1})^{-1}-(t+\sqrt{t^2-1})^{-1}\end{pmatrix}\begin{pmatrix}2\sqrt{t^2-1}\end{pmatrix}^{-1}\\ &={\frac 1{2\sqrt{t^2-1}(t-\sqrt{t^2-1})}}-{\frac 1{2\sqrt{t^2-1}(t+\sqrt{t^2-1})}}\\ &={\frac 1{t^2-(t^2-1)}}\\ &=1\\ U_1(t)&=\begin{pmatrix}(t-\sqrt{t^2-1})^{-2}-(t+\sqrt{t^2-1})^{-2}\end{pmatrix}\begin{pmatrix}2\sqrt{t^2-1}\end{pmatrix}^{-1}\\ &={\frac 1{2\sqrt{t^2-1}(2t^2-1-2\sqrt{t^4-t^2})}}-{\frac 1{2\sqrt{t^2-1}(2t^2-1+2\sqrt{t^4-t^2})}}\\ &=2t\\ U_3(t)&=\begin{pmatrix}(t-\sqrt{t^2-1})^{-3}-(t+\sqrt{t^2-1})^{-3}\end{pmatrix}\begin{pmatrix}2\sqrt{t^2-1}\end{pmatrix}^{-1}\\ &=\begin{pmatrix}(t-\sqrt{t^2-1})^{-2}+1+(t+\sqrt{t^2-1})^{-2}\end{pmatrix}\\ &=4t^2-1\end{align*}$

c) In part a) above we have found the power series by looking at power series of

$\begin{align*}\begin{pmatrix}1-{\frac z{\omega_-}}\end{pmatrix}^{-1}, \begin{pmatrix}1-{\frac z{\omega_+}}\end{pmatrix}^{-1}\end{align*}$

These power series converge if and only if $\begin{align*}|z|<|w_-|\end{align*}$ , respectively, $\begin{align*}|z|<|w_+|\end{align*}$ . Therefore, the power series for $f_t(z)$ converges if and only if $|z|<\min\{|w_-|,|w_+|\}$ . Thus, the radius of convergence is

$\min\{|w_-|,|w_+|\}=\min\{|t-\sqrt{t^2-1}|,|t+\sqrt{t^2-1}|\}$