Statement (I): f(z) = sinz can be expanded at $\mathrm{z}=\pi /4$ by using Laurent series

Statement (I): $f\left(z\right)=\frac{1}{z+4}$ can be expanded by Taylor’s series in a valid region, 1 < |z| < 3

Which of the statements are true

This problem assesses understanding of complex series expansions, specifically distinguishing between Taylor series and Laurent series based on the function's analyticity and the specified region of expansion.

Series Expansion Analysis: Statement I Explained

Statement (I): $f(z)=\sin z$ can be expanded at $z=\pi /4$ by using Laurent series.

- **Understanding $f(z)=\sin z$:** The function $f(z)=\sin z$ is an entire function. This means it is analytic (differentiable at every point) everywhere in the complex plane.
- **Analyticity at $z=\pi /4$:** Since $f(z)=\sin z$ is analytic at $z=\pi /4$, its series expansion around this point will be a Taylor series.
- Taylor Series vs. Laurent Series: A Taylor series is used to expand a function $f(z)$ around a point $z_0$ where the function is analytic. It converges in the largest open disk centered at $z_0$ where $f(z)$ is analytic. A Taylor series only contains non-negative powers of $(z-z_0)$. A Laurent series is a more general expansion used when a function has an isolated singularity at $z_0$, or when expanding in an annular region around a singularity. It includes both non-negative powers (analytic part) and negative powers (principal part) of $(z-z_0)$.
- Interpreting the Statement: While a Taylor series is mathematically a special case of a Laurent series (where all coefficients of negative powers are zero), typically, when a function is analytic at the point of expansion, the appropriate and more precise description is a Taylor series. Stating "by using Laurent series" in this context can be considered misleading as it often implies the presence of a principal part or a non-analytic nature at the center, which is not true for $\sin z$ at $z=\pi /4$. Therefore, for the purpose of distinguishing these series in typical problems, Statement (I) is considered false.

Laurent Series Expansion: Statement II Examined

Statement (II): $f\left(z\right)=\frac{1}{z+4}$ can be expanded by Taylor’s series in a valid region, $1<|z|<3$.

- Function's Singularity: The function $f(z)=\frac{1}{z+4}$ has a singularity at $z=-4$.
- Nature of Taylor Series Region: A Taylor series expansion, centered at a point $z_0$, is always valid in an open disk of the form $|z-z_0|<R$. This disk extends up to the nearest singularity of the function.
- Given Region of Expansion: The region specified is $1<|z|<3$. This region is an annulus (a ring-shaped region), not an open disk.
- Series Type for Annular Region: Expansions of functions in annular regions are characteristic of a Laurent series, not a Taylor series. Even though the function $f(z)=\frac{1}{z+4}$ is analytic within this particular annulus (because its singularity at $z=-4$ lies outside this region, as $|-4|=4$, which is not between 1 and 3), the key point is that a Taylor series' region of convergence is fundamentally an open disk.
- Conclusion for Statement II: Since the region $1<|z|<3$ is an annulus and not an open disk, it cannot be the region of convergence for a Taylor series. Therefore, Statement (II) is false.

Summary of Statements' Validity

Based on the detailed analysis:

- Statement (I) is false because while a Taylor series is a specific case of a Laurent series, the appropriate and precise term for expanding an analytic function at an analytic point is a Taylor series.
- Statement (II) is false because a Taylor series converges in an open disk, whereas the given region $1<|z|<3$ is an annulus. Laurent series are used for expansions in annular regions.

Therefore, neither of the statements is true.