Exercise · ZW-LAURENT-SERIES-010 · ★★★ Advanced
Which of these functions does NOT admit a Laurent series on the annulus \(0<|z|<1\)?
Choose one option.
Show answer
- B. \(\dfrac1{z-\frac12}\)
Explanation
A Laurent series on an annulus requires \(f\) to be holomorphic on all of it. \(\frac1{z-1/2}\) has a pole at \(z=\frac12\), inside \(0<|z|<1\). The others are holomorphic there.
Review the concept: Laurent series
The Laurent series of a function \(f\) holomorphic on an annulus \(r<|z-z_0|<R\) is its expansion in integer powers, positive and negative, of \(z-z_0\), with coefficients \(a_n\).
\[f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n\]
More practice problems: laurent series
- ZW-LAURENT-SERIES-001 What is the coefficient of z^-3 in the Laurent series of e^1/z on 0<|z|?
- ZW-LAURENT-SERIES-002 Find the coefficient of z^-1 in the Laurent expansion of z^2sin (1/z) on 0<|z|.
- ZW-LAURENT-SERIES-003 On 0<|z|<1, how many terms with a negative power does the Laurent series of 1z^2(1-z) have?
- ZW-LAURENT-SERIES-004 What is the Laurent series of 1z-1 on the region |z|>1?
- ZW-LAURENT-SERIES-005 Compute oint_|z|=1e^3/z dz, with the circle traversed once counterclockwise.
- ZW-LAURENT-SERIES-006 If f is holomorphic on the whole disc |z-z_0|<R, what does its Laurent series on 0<|z-z_0|<R look l…
- ZW-LAURENT-SERIES-007 Let f(z)=1(z-1)(z-2). On the annulus 1<|z|<2, what is the coefficient a_0?
- ZW-LAURENT-SERIES-008 Which integral gives the coefficient a_-2 of the Laurent series of f about z_0 (over a circle insid…