Consider the function f defined by f(z) = $\frac{1}{1-\mathrm{z}-{\mathrm{z}}^2}$ for z ∈ ℂ such that 1 − z − z² ≠ 0. Which of the following statements is true?

Concept:

An entire function is a complex-valued function that is a complex differential in a neighborhood of each point in a domain in a complex coordinate space

Explanation:

f(z) = $\frac{1}{1-\mathrm{z}-{\mathrm{z}}^2}$ for z ∈ ℂ 

singularities of f(z) is given by

1 - z - z² = 0 ⇒ z = $\frac{-1\pm \sqrt{1+4}}{2}$ = $\frac{-1\pm \sqrt{5}}{2}$

So f(z) has a pole at z = $\frac{-1\pm \sqrt{5}}{2}$ 

So options (1) and (2) both false

f has a Taylor series expansion so

f(z) = ${\displaystyle \sum _{\mathrm{n}=0}^{\mathrm{\infty }}}$ anzn = f(0) + $\frac{f^{\prime }(0)}{1!}$z + .... 

So a0 = f(0) = 1

and a1 = $\frac{f^{\prime }(0)}{1!}$

Now, f'(z) = -$\frac{1}{(1-\mathrm{z}-{\mathrm{z}}^2{)}^2}$(-1 - 2z)

So f'(0) = 1

So option (4) is correct and (3) false