find-laurent-series-f-z-1-z-2-z-1-0-lt-z-1-lt-1-

Question Number 148724 by Sozan last updated on 30/Jul/21

f i n d l a u r e n t s e r i e s f (z) = \frac{1}{z^{2} - z + 1}, 0 <∣ z - 1 ∣< 1

Answered by mathmax by abdo last updated on 30/Jul/21

the way of this kind is to use changement z - 1 = y \Rightarrow

f (z) = φ (y) = \frac{1}{{(y + 1)}^{2} - (y + 1) + 1} = \frac{1}{y^{2} + 2 y + 1 - y}

= \frac{1}{y^{2} + y + 1} roots of y^{2} + y + 1 = 0

Δ = 1 - 4 = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{2 i π}{3}} and z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{2 i π}{3}}

\Rightarrow φ (y) = \frac{1}{(y - z_{1}) (y - z_{2})} = \frac{1}{z_{1} - z_{2}} (\frac{1}{y - z_{1}} - \frac{1}{y - z_{2}})

= \frac{1}{2 i . \frac{\sqrt{3}}{2}} (\frac{1}{y - z_{1}} - \frac{1}{y - z_{2}}) we have ∣ \frac{y}{z_{1}} ∣=∣ y ∣< 1

∣ \frac{y}{z_{2}} ∣=∣ y ∣< 1 \Rightarrow φ (y) = \frac{1}{i \sqrt{3}} {\frac{1}{z_{1} (\frac{y}{z_{1}} - 1)} - \frac{1}{z_{2} (\frac{y}{z_{2}} - 1)}}

= \frac{1}{i \sqrt{3}} (\frac{1}{z_{2}} \times \frac{1}{1 - \frac{y}{z_{2}}} - \frac{1}{z_{1}} \times \frac{1}{1 - \frac{y}{z_{1}}})

= \frac{e^{\frac{2 i π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} \frac{y^{n}}{z_{2}^{n}} - \frac{e^{- \frac{2 i π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} \frac{y^{n}}{z_{1}^{n}}

= \frac{1}{i \sqrt{3}} e^{\frac{2 i π}{3}} \sum_{n = 0}^{\infty} e^{\frac{i 2 n π}{3}} y^{n} - \frac{1}{i \sqrt{3}} e^{\frac{- 2 i π}{3}} \sum_{n = 0}^{\infty} e^{- \frac{i 2 n π}{3}} y^{n}

= \frac{1}{i \sqrt{3}} \sum_{n = 0}^{\infty} e^{\frac{i (2 n + 1) π}{3}} {(z - 1)}^{n} - \frac{1}{i \sqrt{3}} \sum_{n = 0}^{\infty} e^{- \frac{i (2 n + 1) π}{3}} {(z - 1)}^{n}

= \frac{1}{i \sqrt{3}} \sum_{n = 0}^{\infty} 2 isin (\frac{(2 n + 1) π}{3}) {(z - 1)}^{n} \Rightarrow

f (z) = \frac{2}{\sqrt{3}} \sum_{n = 0}^{\infty} \sin (\frac{(2 n + 1) π}{3}) {(z - 1)}^{n}